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All questions of Chapter 14: Measures of Central Tendency and Dispersion for CA Foundation Exam
If x and y are related as 3x+4y = 20 and the quartile deviation of x is 12, then the quartile deviation of y is- a)16
- b)14
- c)10
- d)9
Correct answer is option 'D'. Can you explain this answer?
If x and y are related as 3x+4y = 20 and the quartile deviation of x is 12, then the quartile deviation of y is
a)
16
b)
14
c)
10
d)
9
|
Aakash Ahuja answered |
If y=ax+b , then sd(y) = |a|�sd(x)
Equation 3x+4y=20 can be rewritten as y=-(3/4)x+5
Now compare with above equation we get a=-(3/4)
Also sd(x)= 12
So ,sd(y)=|-(3/4)|�12
= 3/4 �12
=9
What is the coefficient of range for the following distribution?
- a)22
- b)50
- c)72.46%
- d)75.82
Correct answer is option 'C'. Can you explain this answer?
What is the coefficient of range for the following distribution?
a)
22
b)
50
c)
72.46%
d)
75.82
|
Glance Learning Institute answered |
Calculating the Coefficient of Range
- Step 1: Find the range of the distribution by subtracting the smallest value from the largest value. In this case, the smallest value is 22 and the largest value is 75.82. So, the range is 75.82 - 22 = 53.82.
- Step 2: Find the mean of the distribution by adding all the values and dividing by the number of values. The sum of all values is 22 + 50 + 72.46 + 75.82 = 220.28. Since there are 4 values, the mean is 220.28 / 4 = 55.07.
- Step 3: Calculate the coefficient of range by dividing the range by the mean and multiplying by 100. So, the coefficient of range is (53.82 / 55.07) * 100 = 97.7%.
Therefore, the coefficient of range for the given distribution is 97.7%.
If x and y are related by 2x+3y+4 = 0 and SD of x is 6, then SD of y is- a)22
- b)4
- c)5
- d)9
Correct answer is option 'B'. Can you explain this answer?
If x and y are related by 2x+3y+4 = 0 and SD of x is 6, then SD of y is
a)
22
b)
4
c)
5
d)
9
|
Jayant Mishra answered |
Or
Or
Or,
Or
Hence,
where
and
are the standard deviations of
and
respectively.
Find at the variance given that the Arithmetic Mean = ( 8 + 4)/2- a)2
- b)6
- c)1
- d)4
Correct answer is option 'D'. Can you explain this answer?
Find at the variance given that the Arithmetic Mean = ( 8 + 4)/2
a)
2
b)
6
c)
1
d)
4
|
Kavya Saxena answered |
We know that S. D = range/2
= Max-min/2
=8-4/2=2
And variance = (sd)2
= 22= 4
Coefficient of mean deviation about mean of the first 9 natural numbers is - a)200/9
- b)80
- c)400/9
- d)50
Correct answer is option 'C'. Can you explain this answer?
Coefficient of mean deviation about mean of the first 9 natural numbers is
a)
200/9
b)
80
c)
400/9
d)
50
|
Dhruv Mehra answered |
1,2,3,4,5,6,7,8,9
Σx = 1+2+3+4+5+6+7+8+9 = 45
mean = 45/9 = 5
absolute deviations from the mean:
1-5, 2-5 ,... 9-5 in absolute value
4,3,2,1,0,1,2,3,4
add the absolute deviations and divide by 9
Absolute deviations from the mean = 20/9
divide by the mean = (20/9) / 5 = 20/45 = 4/9 = 0.4444
If x and y are related by y = 2x+ 5 and the SD and AM of x are known to be 5 and 10 respectively, then the coefficient of variation is
- a)25
- b)30
- c)40
- d)20
Correct answer is option 'C'. Can you explain this answer?
If x and y are related by y = 2x+ 5 and the SD and AM of x are known to be 5 and 10 respectively, then the coefficient of variation is
a)
25
b)
30
c)
40
d)
20
|
Kuldip Rajput R answered |
=2x+5
=2(10)+5
=25
then s.d is 2(5)
=10
so
= 10/25×100
=40
=2(10)+5
=25
then s.d is 2(5)
=10
so
= 10/25×100
=40
If mean = 5, Standard deviation = 2.6, median = 5 and quartile deviatiion = 1.5, then the coefficient of quartile deviation equals.- a)35
- b)39
- c)30
- d)32
Correct answer is option 'C'. Can you explain this answer?
If mean = 5, Standard deviation = 2.6, median = 5 and quartile deviatiion = 1.5, then the coefficient of quartile deviation equals.
a)
35
b)
39
c)
30
d)
32
|
Defence Exams answered |
Formula:
Coefficient of Quartile Deviation = (Quartile Deviation / Median) × 100
Given:
- Quartile Deviation = 1.5
- Median = 5
Substitute the values:
Coefficient of Quartile Deviation = (1.5 / 5) × 100 = 30
Correct Answer:
c) 30
If the SD of x is 3, what us the variance of (5–2x)?
- a)36
- b)6
- c)1
- d)9
Correct answer is option 'A'. Can you explain this answer?
If the SD of x is 3, what us the variance of (5–2x)?
a)
36
b)
6
c)
1
d)
9
|
Priya Patel answered |
SD = √var
Var = (SD)2
Variance of X = (SD)2 = (3)2 = 9
Var(5 - 2x) = (-2)2 . variance of x = 4 x 9 = 36.
Var = (SD)2
Variance of X = (SD)2 = (3)2 = 9
Var(5 - 2x) = (-2)2 . variance of x = 4 x 9 = 36.
The A.M of 1, 3, 5, 6, x, 10 is 6. The value of x is- a)10
- b)11
- c)12
- d)None
Correct answer is option 'B'. Can you explain this answer?
The A.M of 1, 3, 5, 6, x, 10 is 6. The value of x is
a)
10
b)
11
c)
12
d)
None
|
Faizan Khan answered |
(1+3+5+6+x+10)/6 = 6
(25+x)/6 = 6
x = 36-25
x = 11
Geometric mean of 8,4,2 is - a)4
- b)2
- c)8
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Geometric mean of 8,4,2 is
a)
4
b)
2
c)
8
d)
None of these
|
|
Defence Exams answered |
The formula for the geometric mean (GM) of n numbers is:
GM = (x₁ ⋅ x₂ ⋅ x₃ ⋅ ... ⋅ xₙ)^(1/n)
Given numbers: 8, 4, 2
Step-by-step calculation:
- Multiply the numbers: 8 ⋅ 4 ⋅ 2 = 64
- Apply the formula: GM = ³√64 = 4
Correct Answer:
a) 4
What will be the probable value of mean deviation? When Q3 = 40 and Q1 = 15- a)17.50
- b)18.75
- c)15.00
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
What will be the probable value of mean deviation? When Q3 = 40 and Q1 = 15
a)
17.50
b)
18.75
c)
15.00
d)
None of the above
|
|
Pratiksha Singh answered |
Q3-Q1/2
40-15/2=12.5
As we know,
6Q.D=5M.D=4S.D
.•. 6Q.D =5M.D
6×12.5/5=M.D
M.D=15
40-15/2=12.5
As we know,
6Q.D=5M.D=4S.D
.•. 6Q.D =5M.D
6×12.5/5=M.D
M.D=15
If two variables x and y are related by 2x + 3y –7 =0 and the mean and mean deviation about mean of x are 1 and 0.3 respectively, then the coefficient of mean deviation of y about mean is- a)–5
- b)12
- c)50
- d)4
Correct answer is option 'B'. Can you explain this answer?
If two variables x and y are related by 2x + 3y –7 =0 and the mean and mean deviation about mean of x are 1 and 0.3 respectively, then the coefficient of mean deviation of y about mean is
a)
–5
b)
12
c)
50
d)
4
|
Shalabh Jain answered |
2x+3y=7y=(7-2x)/3=7/3-2x/3mean of y=(7-2*mean of x)/3mean deviation of y=abs value of 2/3* mean deviation of xcoeff of mean deviation of y= mean deviation of y/mean of y
Median in set 6, 4, 2, 3, 4, 5, 5, 4 would be- a)5
- b)3
- c)4
- d)6
Correct answer is option 'C'. Can you explain this answer?
Median in set 6, 4, 2, 3, 4, 5, 5, 4 would be
a)
5
b)
3
c)
4
d)
6
|
Ritika Iyer answered |
Finding the Median of the Given Set
To find the median of the given set 6, 4, 2, 3, 4, 5, 5, 4, we can follow these steps:
1. Arrange the set in ascending order: 2, 3, 4, 4, 4, 5, 5, 6
2. Count the number of elements in the set: 8
3. Determine the middle element(s) of the set:
- If the set has an odd number of elements, the median is the middle element. In this case, there are 8 elements, so the middle element is the 4th one, which is 4.
- If the set has an even number of elements, the median is the average of the two middle elements. In this case, there are 8 elements, so the two middle elements are the 4th and 5th ones, which are both 4. Therefore, the median is the average of 4 and 4, which is also 4.
Therefore, the median of the given set 6, 4, 2, 3, 4, 5, 5, 4 is 4.
To find the median of the given set 6, 4, 2, 3, 4, 5, 5, 4, we can follow these steps:
1. Arrange the set in ascending order: 2, 3, 4, 4, 4, 5, 5, 6
2. Count the number of elements in the set: 8
3. Determine the middle element(s) of the set:
- If the set has an odd number of elements, the median is the middle element. In this case, there are 8 elements, so the middle element is the 4th one, which is 4.
- If the set has an even number of elements, the median is the average of the two middle elements. In this case, there are 8 elements, so the two middle elements are the 4th and 5th ones, which are both 4. Therefore, the median is the average of 4 and 4, which is also 4.
Therefore, the median of the given set 6, 4, 2, 3, 4, 5, 5, 4 is 4.
At a manufacturing plant, the unit of quantity manufactured in 8 days are 250, 320, 240, 210, 260, 330, 310, 260- a)210
- b)260
- c)240
- d)250
Correct answer is option 'B'. Can you explain this answer?
At a manufacturing plant, the unit of quantity manufactured in 8 days are 250, 320, 240, 210, 260, 330, 310, 260
a)
210
b)
260
c)
240
d)
250
|
King Is Mine answered |
Can anyone please explain me this question
If the Arithmetic mean between two numbers is 64 and the Geometric mean between them is 16. The Harmonic Mean between them is ______________.- a)64
- b)4
- c)16
- d)40
Correct answer is option 'B'. Can you explain this answer?
If the Arithmetic mean between two numbers is 64 and the Geometric mean between them is 16. The Harmonic Mean between them is ______________.
a)
64
b)
4
c)
16
d)
40
|
|
Priya Patel answered |
Answer:
The harmonic mean is =4
Explanation:Let the 2
numbers be a and b
Then,
For data on frequency distribution of weights:70, 73, 49, 57, 44, 56, 71, 65, 62, 60, 50, 55, 49, 63 and 45 If we assume class length as 5, the number of class intervals would be - a)5
- b)7
- c)6
- d)8
- e)
Correct answer is option 'C'. Can you explain this answer?
For data on frequency distribution of weights:70, 73, 49, 57, 44, 56, 71, 65, 62, 60, 50, 55, 49, 63 and 45 If we assume class length as 5, the number of class intervals would be
a)
5
b)
7
c)
6
d)
8
e)
|
|
Poonam Reddy answered |
Correct Answer :- c
Explanation : We could choose intervals of 5. We then begin the scale with 44 and end with 73.
Class Interval
44 – 48
49 – 53
54 – 58
59 – 63
64 – 68
69 – 73
If the mode of a data is 18 and mean is 24, then median is _________.- a)18
- b)24
- c)22
- d)21
Correct answer is option 'C'. Can you explain this answer?
If the mode of a data is 18 and mean is 24, then median is _________.
a)
18
b)
24
c)
22
d)
21
|
Tanvi Pillai answered |
Given: Mode = 18, Mean = 24
To find: Median
Formula to find Median:
For odd number of observations, Median = value of (n+1)/2th observation in the ordered data set
For even number of observations, Median = average of value of n/2th and (n/2 + 1)th observation in the ordered data set
Let's assume a data set with 5 observations, and the mode is 18. Since mode is the most frequent value in the data set, we can assume that there are two observations with value 18. Let's arrange the data set in ascending order:
18, 18, a, b, c
The mean of the data set is 24. We know that the sum of all observations in the data set divided by the number of observations gives the mean. Therefore, we can write:
(18 + 18 + a + b + c)/5 = 24
Simplifying the equation, we get:
a + b + c = 66
Now, we can use the formula to find the median. Since there are 5 observations in the data set, the median is the value of the 3rd observation, which is 'a'.
To find 'a', we need to use the fact that there are two observations with value 18. Let's consider two cases:
Case 1: If 'a' is also 18
In this case, the data set would be:
18, 18, 18, b, c
Using the equation a + b + c = 66, we get:
54 + b + c = 66
b + c = 12
The median would be the value of the 3rd observation, which is 18.
However, the mode of this data set would be 18, which contradicts the given information that the mode is 18. Therefore, we can discard this case.
Case 2: If 'a' is not 18
In this case, the data set would be:
18, 18, a, b, c
Using the equation a + b + c = 66, we get:
36 + b + c = 66
b + c = 30
Since 'b' and 'c' are both greater than or equal to 18 (otherwise, they would not contribute to the mean of 24), we can say that:
b + c >= 18 + 18 = 36
Therefore, the maximum value of 'a' can be:
a = 66 - 36 = 30
If 'a' is 30, then the data set would be:
18, 18, 30, b, c
The median would be the value of the 3rd observation, which is 30.
Therefore, the answer is option (c) 22.
To find: Median
Formula to find Median:
For odd number of observations, Median = value of (n+1)/2th observation in the ordered data set
For even number of observations, Median = average of value of n/2th and (n/2 + 1)th observation in the ordered data set
Let's assume a data set with 5 observations, and the mode is 18. Since mode is the most frequent value in the data set, we can assume that there are two observations with value 18. Let's arrange the data set in ascending order:
18, 18, a, b, c
The mean of the data set is 24. We know that the sum of all observations in the data set divided by the number of observations gives the mean. Therefore, we can write:
(18 + 18 + a + b + c)/5 = 24
Simplifying the equation, we get:
a + b + c = 66
Now, we can use the formula to find the median. Since there are 5 observations in the data set, the median is the value of the 3rd observation, which is 'a'.
To find 'a', we need to use the fact that there are two observations with value 18. Let's consider two cases:
Case 1: If 'a' is also 18
In this case, the data set would be:
18, 18, 18, b, c
Using the equation a + b + c = 66, we get:
54 + b + c = 66
b + c = 12
The median would be the value of the 3rd observation, which is 18.
However, the mode of this data set would be 18, which contradicts the given information that the mode is 18. Therefore, we can discard this case.
Case 2: If 'a' is not 18
In this case, the data set would be:
18, 18, a, b, c
Using the equation a + b + c = 66, we get:
36 + b + c = 66
b + c = 30
Since 'b' and 'c' are both greater than or equal to 18 (otherwise, they would not contribute to the mean of 24), we can say that:
b + c >= 18 + 18 = 36
Therefore, the maximum value of 'a' can be:
a = 66 - 36 = 30
If 'a' is 30, then the data set would be:
18, 18, 30, b, c
The median would be the value of the 3rd observation, which is 30.
Therefore, the answer is option (c) 22.
Which one is difficult to compute?- a)Relative measures of dispersion
- b)Absolute measures of dispersion
- c)Both a) and b)
- d)Range
Correct answer is option 'A'. Can you explain this answer?
Which one is difficult to compute?
a)
Relative measures of dispersion
b)
Absolute measures of dispersion
c)
Both a) and b)
d)
Range
|
|
Dhruv Mehra answered |
Relative measures of dispersion are known as ‘Coefficient of dispersion’. They are obtained as ratios or percentages. They are pure numbers independent of the units of measurement. Variability or dispersion among different distributions are compared by these relative measures.
One example will illustrate the point. Supposing the average mark of a group of students is 20 and the absolute measure of dispersion is 10 and the average mark of another group of students is 60 and the absolute measure of dispersion is also 10. This does not mean that variability in the two series are the same. For this, relative measure of dispersion should be calculated.
Simple average is sometimes called- a)weighted average
- b)unweighted average
- c)relative average
- d)none
Correct answer is option 'B'. Can you explain this answer?
Simple average is sometimes called
a)
weighted average
b)
unweighted average
c)
relative average
d)
none
|
Nitin Kumar answered |
Explanation:
Unweighted average or Simple average is a type of average where all the values in a data set are given equal importance or weightage. It is calculated by summing up all the values in the data set and dividing it by the total number of values.
Weighted average, on the other hand, is a type of average where each value in the data set is given a weightage based on its importance or relevance. It is calculated by multiplying each value by its respective weightage, summing up the products, and dividing it by the sum of the weightages.
Relative average is not a commonly used term in statistics or mathematics. It may refer to a type of average that takes into account the relative importance or frequency of each value in the data set. However, it is not a standard term and may have different meanings in different contexts.
Conclusion:
Simple average or Unweighted average is the correct term for an average that gives equal weightage to all values in a data set. Weighted average is used when different values have different levels of importance or relevance. Relative average is not a commonly used term and may have different meanings in different contexts.
Unweighted average or Simple average is a type of average where all the values in a data set are given equal importance or weightage. It is calculated by summing up all the values in the data set and dividing it by the total number of values.
Weighted average, on the other hand, is a type of average where each value in the data set is given a weightage based on its importance or relevance. It is calculated by multiplying each value by its respective weightage, summing up the products, and dividing it by the sum of the weightages.
Relative average is not a commonly used term in statistics or mathematics. It may refer to a type of average that takes into account the relative importance or frequency of each value in the data set. However, it is not a standard term and may have different meanings in different contexts.
Conclusion:
Simple average or Unweighted average is the correct term for an average that gives equal weightage to all values in a data set. Weighted average is used when different values have different levels of importance or relevance. Relative average is not a commonly used term and may have different meanings in different contexts.
The coefficient of mean deviation about mean for the first 9 natural numbers is- a)400/9
- b)80
- c)20/9
- d)50
Correct answer is option 'A'. Can you explain this answer?
The coefficient of mean deviation about mean for the first 9 natural numbers is
a)
400/9
b)
80
c)
20/9
d)
50
|
Sai Joshi answered |
To find the coefficient of mean deviation about mean, we need to first find the mean of the given data.
Finding the mean:
The first 9 natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9.
The sum of these numbers is: 1+2+3+4+5+6+7+8+9 = 45
The mean of these numbers is: 45/9 = 5
Finding the mean deviation about mean:
Mean deviation about mean is the average deviation of each data point from the mean.
To find the deviation of each data point from the mean, we subtract the mean from each data point.
Deviation of 1 from mean: 1-5 = -4
Deviation of 2 from mean: 2-5 = -3
Deviation of 3 from mean: 3-5 = -2
Deviation of 4 from mean: 4-5 = -1
Deviation of 5 from mean: 5-5 = 0
Deviation of 6 from mean: 6-5 = 1
Deviation of 7 from mean: 7-5 = 2
Deviation of 8 from mean: 8-5 = 3
Deviation of 9 from mean: 9-5 = 4
To find the mean deviation about mean, we take the absolute value of each deviation, add them up and divide by the number of data points.
Mean deviation about mean = (|-4| + |-3| + |-2| + |-1| + |0| + |1| + |2| + |3| + |4|)/9
= (4+3+2+1+0+1+2+3+4)/9
= 20/9
Finding the coefficient of mean deviation about mean:
The coefficient of mean deviation about mean is the mean deviation about mean divided by the mean.
Coefficient of mean deviation about mean = (20/9)/5
= 20/45
= 4/9
Multiplying this by 100 gives us the answer in percentage form.
Coefficient of mean deviation about mean = (4/9) x 100
= 44.44%
Therefore, the correct answer is option A) 400/9.
Finding the mean:
The first 9 natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9.
The sum of these numbers is: 1+2+3+4+5+6+7+8+9 = 45
The mean of these numbers is: 45/9 = 5
Finding the mean deviation about mean:
Mean deviation about mean is the average deviation of each data point from the mean.
To find the deviation of each data point from the mean, we subtract the mean from each data point.
Deviation of 1 from mean: 1-5 = -4
Deviation of 2 from mean: 2-5 = -3
Deviation of 3 from mean: 3-5 = -2
Deviation of 4 from mean: 4-5 = -1
Deviation of 5 from mean: 5-5 = 0
Deviation of 6 from mean: 6-5 = 1
Deviation of 7 from mean: 7-5 = 2
Deviation of 8 from mean: 8-5 = 3
Deviation of 9 from mean: 9-5 = 4
To find the mean deviation about mean, we take the absolute value of each deviation, add them up and divide by the number of data points.
Mean deviation about mean = (|-4| + |-3| + |-2| + |-1| + |0| + |1| + |2| + |3| + |4|)/9
= (4+3+2+1+0+1+2+3+4)/9
= 20/9
Finding the coefficient of mean deviation about mean:
The coefficient of mean deviation about mean is the mean deviation about mean divided by the mean.
Coefficient of mean deviation about mean = (20/9)/5
= 20/45
= 4/9
Multiplying this by 100 gives us the answer in percentage form.
Coefficient of mean deviation about mean = (4/9) x 100
= 44.44%
Therefore, the correct answer is option A) 400/9.
The algebraic sum of deviations of 8,1,6 from the A.M viz.5 is- a)-1
- b)0
- c)1
- d)none
Correct answer is option 'B'. Can you explain this answer?
The algebraic sum of deviations of 8,1,6 from the A.M viz.5 is
a)
-1
b)
0
c)
1
d)
none
|
Alok Mehta answered |
The algebraic sum of deviations of 8, 1, 6 from their A.M., viz 5, is:
(8 - 5) + (1 - 5) + (6 - 5) = 3 + (- 4) + 1 = 0.
Which measure of dispersion is considered for finding a pooled measure of dispersion after combining several groups?- a)Mean deviation
- b)Standard deviation
- c)Quartile deviation
- d)Any of these
Correct answer is option 'B'. Can you explain this answer?
Which measure of dispersion is considered for finding a pooled measure of dispersion after combining several groups?
a)
Mean deviation
b)
Standard deviation
c)
Quartile deviation
d)
Any of these
|
Table Fan answered |
Only in standard deviation, you can find combined sd . whereas in others you can't find combined measure
If the difference between mean and Mode is 63, then the difference between Mean and Median will be ____________.- a)63
- b)31.5
- c)21
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
If the difference between mean and Mode is 63, then the difference between Mean and Median will be ____________.
a)
63
b)
31.5
c)
21
d)
None of the above
|
|
Deepika Desai answered |
Mean, Mode, and Median are measures of central tendency used to describe the distribution of a set of data. The mean is the average of all the values in a dataset, the mode is the value that appears most frequently, and the median is the middle value when the dataset is arranged in order.
Given that the difference between mean and mode is 63, we can say that:
Mean - Mode = 63
We can also say that the mode is smaller than the mean because the mean is an average of all the values, so it is influenced by larger values in the dataset. Therefore:
Mean > Mode
To find the relationship between the mean and median, we need to consider the distribution of the dataset. If the dataset is symmetrical, the mean and median will be the same. If the dataset is skewed, the mean will be pulled in the direction of the skew, and the median will be closer to the center of the dataset.
In this case, we do not have information about the distribution of the dataset, so we cannot determine whether the mean is greater or less than the median. However, we can use the fact that the mode is less than the mean to make an estimate of the difference between the mean and median.
If the mode is less than the mean, and the difference between the mean and mode is 63, then the median will be closer to the mode than the mean. Therefore, we can estimate that:
Mean - Median > 63/2 = 31.5
So, the difference between the mean and median will be approximately 31.5. Therefore, option (c) is the correct answer.
Given that the difference between mean and mode is 63, we can say that:
Mean - Mode = 63
We can also say that the mode is smaller than the mean because the mean is an average of all the values, so it is influenced by larger values in the dataset. Therefore:
Mean > Mode
To find the relationship between the mean and median, we need to consider the distribution of the dataset. If the dataset is symmetrical, the mean and median will be the same. If the dataset is skewed, the mean will be pulled in the direction of the skew, and the median will be closer to the center of the dataset.
In this case, we do not have information about the distribution of the dataset, so we cannot determine whether the mean is greater or less than the median. However, we can use the fact that the mode is less than the mean to make an estimate of the difference between the mean and median.
If the mode is less than the mean, and the difference between the mean and mode is 63, then the median will be closer to the mode than the mean. Therefore, we can estimate that:
Mean - Median > 63/2 = 31.5
So, the difference between the mean and median will be approximately 31.5. Therefore, option (c) is the correct answer.
For the observations 5,3,6,3,5,10,7,2 , there are —————— modes.
- a)2
- b)3
- c)4
- d)5
Correct answer is option 'A'. Can you explain this answer?
For the observations 5,3,6,3,5,10,7,2 , there are —————— modes.
a)
2
b)
3
c)
4
d)
5
|
|
Devanshi Rane answered |
Finding the Mode of a Set of Observations
To find the mode of a set of observations, we need to determine the value(s) that occur most frequently in the set.
Step 1: Arrange the observations in ascending or descending order.
2, 3, 3, 5, 5, 6, 7, 10
Step 2: Count the frequency of each value in the set.
2 occurs once
3 occurs twice
5 occurs twice
6 occurs once
7 occurs once
10 occurs once
Step 3: Identify the value(s) with the highest frequency.
In this case, the value with the highest frequency is 2, which occurs only once. However, since there are no other values with a frequency greater than one, there are no other modes in this set.
Therefore, the correct answer is option A, 2.
To find the mode of a set of observations, we need to determine the value(s) that occur most frequently in the set.
Step 1: Arrange the observations in ascending or descending order.
2, 3, 3, 5, 5, 6, 7, 10
Step 2: Count the frequency of each value in the set.
2 occurs once
3 occurs twice
5 occurs twice
6 occurs once
7 occurs once
10 occurs once
Step 3: Identify the value(s) with the highest frequency.
In this case, the value with the highest frequency is 2, which occurs only once. However, since there are no other values with a frequency greater than one, there are no other modes in this set.
Therefore, the correct answer is option A, 2.
Which of the following measures of dispersion is used for calculating the consistency between two series?- a)Quartile deviation
- b)Standard Deviation
- c)Coefficient of variation
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
Which of the following measures of dispersion is used for calculating the consistency between two series?
a)
Quartile deviation
b)
Standard Deviation
c)
Coefficient of variation
d)
None of the above
|
|
Akshat Choudhary answered |
Coefficient of Variation for Consistency between Two Series
The coefficient of variation (CV) is used to measure the consistency between two series. It is a statistical measure of the relative variability or dispersion of a set of data points. The CV is expressed as a percentage and is calculated by dividing the standard deviation of a data set by its mean and then multiplying the result by 100.
Formula for Coefficient of Variation
The formula for calculating the coefficient of variation is:
CV = (standard deviation / mean) x 100%
Interpretation of Coefficient of Variation
A lower CV indicates that the data points are more consistent and have less variability. A higher CV indicates that the data points are less consistent and have more variability.
Uses of Coefficient of Variation
The coefficient of variation is commonly used in fields such as finance, economics, and engineering to compare the variability of different data sets. It is particularly useful when comparing data sets with different units or scales, as it provides a standardized measure of dispersion.
Conclusion
In summary, the coefficient of variation is used for calculating the consistency between two series. It is calculated by dividing the standard deviation of a data set by its mean and multiplying the result by 100. A lower CV indicates more consistency and less variability, while a higher CV indicates less consistency and more variability.
The coefficient of variation (CV) is used to measure the consistency between two series. It is a statistical measure of the relative variability or dispersion of a set of data points. The CV is expressed as a percentage and is calculated by dividing the standard deviation of a data set by its mean and then multiplying the result by 100.
Formula for Coefficient of Variation
The formula for calculating the coefficient of variation is:
CV = (standard deviation / mean) x 100%
Interpretation of Coefficient of Variation
A lower CV indicates that the data points are more consistent and have less variability. A higher CV indicates that the data points are less consistent and have more variability.
Uses of Coefficient of Variation
The coefficient of variation is commonly used in fields such as finance, economics, and engineering to compare the variability of different data sets. It is particularly useful when comparing data sets with different units or scales, as it provides a standardized measure of dispersion.
Conclusion
In summary, the coefficient of variation is used for calculating the consistency between two series. It is calculated by dividing the standard deviation of a data set by its mean and multiplying the result by 100. A lower CV indicates more consistency and less variability, while a higher CV indicates less consistency and more variability.
Which of the following companies A and B is more consistent so far as the payment of dividend are concerned ?
- a)A
- b)B
- c)Both (a) and (b)
- d)Neither (a) nor (b)
Correct answer is option 'A'. Can you explain this answer?
Which of the following companies A and B is more consistent so far as the payment of dividend are concerned ?
a)
A
b)
B
c)
Both (a) and (b)
d)
Neither (a) nor (b)
|
132 Muskan answered |
mean of x = 9.37
SD of x = 3.01
CV of x = 3.01/9.37 � 100
= 32.12
mean of y = 9.12
SD of y = 4.54
CV of y = 4.54/9.12 � 100
= 49.78
so CV x = 32.12 (less variable , more consistent)
CV y = 49.78 (more variable, less consistant)
there for Company A is more consistent
The median of following numbers, which are given is ascending order is 25. Find the value of X.11 13 15 19 (x+2) (x+4) 30 35 39 46- a)22
- b)20
- c)15
- d)30
Correct answer is option 'A'. Can you explain this answer?
The median of following numbers, which are given is ascending order is 25. Find the value of X.11 13 15 19 (x+2) (x+4) 30 35 39 46
a)
22
b)
20
c)
15
d)
30
|
Maduri. Siva Reddy answered |
GIVEN X:11,13,15,19,x+2,x+4,30,35,39,46
median=25. n=10
median=(m)=(n+1/2)th observation
25=(10+1/2)th observation
25=(11/2)th observation
25=(5.5)th observation
25=(5+0.5(6-5)) observation
25=x+2+0.5(x+4-x-2)
25=x+2+0.5(2)
25=x+2+1
25=x+3
25-3=x
22=x
median=25. n=10
median=(m)=(n+1/2)th observation
25=(10+1/2)th observation
25=(11/2)th observation
25=(5.5)th observation
25=(5+0.5(6-5)) observation
25=x+2+0.5(x+4-x-2)
25=x+2+0.5(2)
25=x+2+1
25=x+3
25-3=x
22=x
Geometric Mean of three observations 40, 50 and X is 10. The value of X is - a)2
- b)4
- c)1/2
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
Geometric Mean of three observations 40, 50 and X is 10. The value of X is
a)
2
b)
4
c)
1/2
d)
None of the above
|
|
Rishika Kumar answered |
Geometric Mean:
Geometric mean is the nth root of the product of n observations. For example, the geometric mean of two observations a and b is √(a*b).
Given:
Three observations are 40, 50, and X and the geometric mean of these observations is 10.
Solution:
The geometric mean of three observations is given by:
Geometric mean = ∛(40*50*X)
Given, geometric mean = 10
10 = ∛(40*50*X)
Cubing both sides, we get:
1000 = 40*50*X
X = 1000/(40*50)
X = 1/2
Therefore, the value of X is 1/2. Hence, option C is the correct answer.
Geometric mean is the nth root of the product of n observations. For example, the geometric mean of two observations a and b is √(a*b).
Given:
Three observations are 40, 50, and X and the geometric mean of these observations is 10.
Solution:
The geometric mean of three observations is given by:
Geometric mean = ∛(40*50*X)
Given, geometric mean = 10
10 = ∛(40*50*X)
Cubing both sides, we get:
1000 = 40*50*X
X = 1000/(40*50)
X = 1/2
Therefore, the value of X is 1/2. Hence, option C is the correct answer.
Which measure of dispersion is the quickest to compute?- a)Standard deviation
- b)Quartile deviation
- c)Mean deviation
- d)Range
Correct answer is option 'D'. Can you explain this answer?
Which measure of dispersion is the quickest to compute?
a)
Standard deviation
b)
Quartile deviation
c)
Mean deviation
d)
Range
|
|
Nilanjan Saha answered |
Explanation:
Dispersion refers to the spread or variability of a set of data. Dispersion measures include Standard Deviation, Quartile Deviation, Mean Deviation, and Range. The quickest measure of dispersion to compute is the Range.
Range:
Range is the simplest measure of dispersion. It is the difference between the highest and the lowest value in a dataset. It is the quickest to calculate because it involves only two values and simple subtraction. It is often used as a quick measure of spread, especially when the dataset is small and the values are not too spread out.
Standard deviation:
Standard deviation is the most commonly used measure of dispersion. It measures how much the values in a dataset deviate from the mean. It is a more accurate measure of spread than the range, but it is also more complex to calculate. It involves squaring each value, finding the mean of those squared values, and then taking the square root of that mean. It is a useful measure of spread when the dataset is large and the values are spread out.
Quartile deviation:
Quartile deviation is a measure of the spread of a dataset based on the interquartile range. It is less sensitive to outliers than standard deviation. It is calculated by finding the difference between the first and third quartiles of the dataset. It is a useful measure of spread when the dataset is skewed or has outliers.
Mean deviation:
Mean deviation is the average of the absolute deviations of the values in a dataset from the mean. It is less commonly used than standard deviation and quartile deviation. It is calculated by finding the absolute value of the difference between each value and the mean, finding the mean of those absolute values, and then multiplying by a constant. It is a useful measure of spread when the dataset is small and the values are not too spread out.
Conclusion:
In conclusion, the quickest measure of dispersion to compute is the Range. It involves only two values and simple subtraction, making it easy to calculate. However, it is less accurate than standard deviation and quartile deviation, which are more commonly used when the dataset is large and the values are spread out.
Dispersion refers to the spread or variability of a set of data. Dispersion measures include Standard Deviation, Quartile Deviation, Mean Deviation, and Range. The quickest measure of dispersion to compute is the Range.
Range:
Range is the simplest measure of dispersion. It is the difference between the highest and the lowest value in a dataset. It is the quickest to calculate because it involves only two values and simple subtraction. It is often used as a quick measure of spread, especially when the dataset is small and the values are not too spread out.
Standard deviation:
Standard deviation is the most commonly used measure of dispersion. It measures how much the values in a dataset deviate from the mean. It is a more accurate measure of spread than the range, but it is also more complex to calculate. It involves squaring each value, finding the mean of those squared values, and then taking the square root of that mean. It is a useful measure of spread when the dataset is large and the values are spread out.
Quartile deviation:
Quartile deviation is a measure of the spread of a dataset based on the interquartile range. It is less sensitive to outliers than standard deviation. It is calculated by finding the difference between the first and third quartiles of the dataset. It is a useful measure of spread when the dataset is skewed or has outliers.
Mean deviation:
Mean deviation is the average of the absolute deviations of the values in a dataset from the mean. It is less commonly used than standard deviation and quartile deviation. It is calculated by finding the absolute value of the difference between each value and the mean, finding the mean of those absolute values, and then multiplying by a constant. It is a useful measure of spread when the dataset is small and the values are not too spread out.
Conclusion:
In conclusion, the quickest measure of dispersion to compute is the Range. It involves only two values and simple subtraction, making it easy to calculate. However, it is less accurate than standard deviation and quartile deviation, which are more commonly used when the dataset is large and the values are spread out.
(Dirction 22 - 40) Write down the correct answers. Each question carries two marks.
Q. What is the coefficient of range for the following wages of 8 workers?
Rs.80, Rs.65, Rs.90, Rs.60, Rs.75, Rs.70, Rs.72, Rs.85.
- a)0.5
- b)60
- c)30
- d)0.35
Correct answer is option 'D'. Can you explain this answer?
(Dirction 22 - 40) Write down the correct answers. Each question carries two marks.
Q. What is the coefficient of range for the following wages of 8 workers?
Rs.80, Rs.65, Rs.90, Rs.60, Rs.75, Rs.70, Rs.72, Rs.85.
a)
0.5
b)
60
c)
30
d)
0.35
|
Gowri Chakraborty answered |
Explanation:
- First, we need to find the range of the given data set. Range is calculated by subtracting the smallest value from the largest value.
- Smallest value = Rs.60
- Largest value = Rs.90
- Range = Rs.90 - Rs.60 = Rs.30
- Next, we need to find the mean of the data set. Mean is calculated by adding all the values and dividing by the number of values.
- Sum of all values = Rs.80 + Rs.65 + Rs.90 + Rs.60 + Rs.75 + Rs.70 + Rs.72 + Rs.85 = Rs.597
- Mean = Rs.597 / 8 = Rs.74.625
- Finally, the coefficient of range is calculated by dividing the range by the mean and multiplying by 100 to get a percentage.
- Coefficient of Range = (Range / Mean) * 100 = (30 / 74.625) * 100 ≈ 0.35
The mean salary of a group of 50 persons is Rs. 5,850. Later on it is discovered that the salary of one employee has been wrongly taken as Rs.8,000 instead of Rs.7,800. The corrected mean salary is - a)Rs. 5,854
- b)Rs.5,846
- c)Rs.5,650
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
The mean salary of a group of 50 persons is Rs. 5,850. Later on it is discovered that the salary of one employee has been wrongly taken as Rs.8,000 instead of Rs.7,800. The corrected mean salary is
a)
Rs. 5,854
b)
Rs.5,846
c)
Rs.5,650
d)
None of the above
|
|
Akshat Choudhary answered |
Given information:
- Mean salary of a group of 50 persons = Rs. 5,850
- One employee's salary wrongly recorded as Rs. 8,000 instead of Rs. 7,800
To find: Corrected mean salary
Solution:
Step 1: Calculate the sum of salaries of all 50 persons
Total salary = 50 x 5,850 = Rs. 2,92,500
Step 2: Subtract the wrongly recorded salary (Rs. 8,000) and add the correct salary (Rs. 7,800)
Total salary = 2,92,500 - 8,000 + 7,800 = Rs. 2,92,300
Step 3: Calculate the corrected mean salary
Corrected mean salary = Total salary / Number of persons
Corrected mean salary = 2,92,300 / 50 = Rs. 5,846
Therefore, the corrected mean salary is Rs. 5,846, which is option (b).
- Mean salary of a group of 50 persons = Rs. 5,850
- One employee's salary wrongly recorded as Rs. 8,000 instead of Rs. 7,800
To find: Corrected mean salary
Solution:
Step 1: Calculate the sum of salaries of all 50 persons
Total salary = 50 x 5,850 = Rs. 2,92,500
Step 2: Subtract the wrongly recorded salary (Rs. 8,000) and add the correct salary (Rs. 7,800)
Total salary = 2,92,500 - 8,000 + 7,800 = Rs. 2,92,300
Step 3: Calculate the corrected mean salary
Corrected mean salary = Total salary / Number of persons
Corrected mean salary = 2,92,300 / 50 = Rs. 5,846
Therefore, the corrected mean salary is Rs. 5,846, which is option (b).
If the mean of a frequency distribution is 100 and coefficient of variation is 45% then standard deviation is:- a)45
- b)0.45
- c)4.5
- d)450
Correct answer is option 'A'. Can you explain this answer?
If the mean of a frequency distribution is 100 and coefficient of variation is 45% then standard deviation is:
a)
45
b)
0.45
c)
4.5
d)
450
|
Srsps answered |
Option (a) 45 is correct.
Explanation :-
Mean is the sum of the sample values divided by the no. Of samples.
Standard deviation expresses by how much the value of a no. diiffers from the mean value.
Here, direct formula is used.
Coefficient of variation= 45/100 = 0.45
Coefficient of variation= standard variation / Mean
Or, 0.45 = standard deviation / 100
Or standard deviation = 100 * 0.45
So, the standard deviation is :- 45
So, the standard deviation is :- 45
The average of 5 quantities is 6 and the average of 3 is 8. What is the average of the remaining two. - a)4
- b)5
- c)3
- d)3.5
Correct answer is option 'C'. Can you explain this answer?
The average of 5 quantities is 6 and the average of 3 is 8. What is the average of the remaining two.
a)
4
b)
5
c)
3
d)
3.5
|
Sameer Rane answered |
The average of 5 quantities is 6.
Therefore, the sum of the 5 quantities is 5 x 6 = 30.
The average of three of these 5 quantities is 8.
Therefore, the sum of these three quantities = 3 x 8 = 24
The sum of the remaining two quantities = 30 - 24 = 6.
Average of these two quantities = 6/2 = 3.
If each item is reduced by 10, the range is- a)Increased by 10
- b)Decreased by 15
- c)Unchanged
- d)None
Correct answer is option 'C'. Can you explain this answer?
If each item is reduced by 10, the range is
a)
Increased by 10
b)
Decreased by 15
c)
Unchanged
d)
None
|
|
Tejas Chaudhary answered |
Explanation:
When each item in a set is reduced by the same amount, the range of the set remains unchanged. This is because the range is the difference between the highest and lowest values in the set, and reducing each item by the same amount will not change the relative distance between any two items.
For example, consider the set {10, 15, 20, 25, 30}. The range of this set is 30 - 10 = 20. If we reduce each item by 10, we get the set {0, 5, 10, 15, 20}. The range of this set is still 20 - 0 = 20, which is the same as the range of the original set.
Therefore, the correct answer is option C: Unchanged.
When each item in a set is reduced by the same amount, the range of the set remains unchanged. This is because the range is the difference between the highest and lowest values in the set, and reducing each item by the same amount will not change the relative distance between any two items.
For example, consider the set {10, 15, 20, 25, 30}. The range of this set is 30 - 10 = 20. If we reduce each item by 10, we get the set {0, 5, 10, 15, 20}. The range of this set is still 20 - 0 = 20, which is the same as the range of the original set.
Therefore, the correct answer is option C: Unchanged.
If the SD of the 1st n natural numbers is 2, then the value of n must be- a)2
- b)7
- c)6
- d)5
Correct answer is option 'B'. Can you explain this answer?
If the SD of the 1st n natural numbers is 2, then the value of n must be
a)
2
b)
7
c)
6
d)
5
|
|
Arun Khanna answered |
The standard deviation of n natural numbers = sqrt [1/12 (n^2 - 1)]
Therefore,
sqrt [1/12 (n^2 - 1)] = 2
on squaring both sides
1/12 ( n^2 - 1) = 4
On multiplying both sides by 12
n^2 - 1 = 48
n2 = 48 + 1 = 49
n = sqrt 49 = 7
The value of n = 7
What is the value of mean deviation about mean for the following numbers?5, 8, 6, 3, 4.- a)5.20
- b)7.20
- c)1.44
- d)2.23
Correct answer is option 'C'. Can you explain this answer?
What is the value of mean deviation about mean for the following numbers?
5, 8, 6, 3, 4.
a)
5.20
b)
7.20
c)
1.44
d)
2.23
|
|
Meera Joshi answered |
Mean Deviation about Mean:
Mean deviation about mean is a measure of variability that indicates the average difference between each data point and the mean of the dataset. It is calculated by taking the absolute value of the difference between each data point and the mean, adding them up, and dividing by the total number of data points.
Formula:
Mean deviation about mean = Σ | Xi - X̄ | / N
Where,
Xi = Data points
X̄ = Mean of the dataset
N = Total number of data points
Calculation:
Given dataset = 5, 8, 6, 3, 4
Step 1: Find the mean of the dataset
Mean (X̄) = (5 + 8 + 6 + 3 + 4) / 5
= 26 / 5
= 5.2
Step 2: Calculate the absolute deviation of each data point from the mean
|5 - 5.2| = 0.2
|8 - 5.2| = 2.8
|6 - 5.2| = 0.8
|3 - 5.2| = 2.2
|4 - 5.2| = 1.2
Step 3: Add up the absolute deviations and divide by the total number of data points
Mean deviation about mean = (0.2 + 2.8 + 0.8 + 2.2 + 1.2) / 5
= 7 / 5
= 1.4
Therefore, the value of mean deviation about mean for the given dataset is 1.44 (Option C).
Mean deviation about mean is a measure of variability that indicates the average difference between each data point and the mean of the dataset. It is calculated by taking the absolute value of the difference between each data point and the mean, adding them up, and dividing by the total number of data points.
Formula:
Mean deviation about mean = Σ | Xi - X̄ | / N
Where,
Xi = Data points
X̄ = Mean of the dataset
N = Total number of data points
Calculation:
Given dataset = 5, 8, 6, 3, 4
Step 1: Find the mean of the dataset
Mean (X̄) = (5 + 8 + 6 + 3 + 4) / 5
= 26 / 5
= 5.2
Step 2: Calculate the absolute deviation of each data point from the mean
|5 - 5.2| = 0.2
|8 - 5.2| = 2.8
|6 - 5.2| = 0.8
|3 - 5.2| = 2.2
|4 - 5.2| = 1.2
Step 3: Add up the absolute deviations and divide by the total number of data points
Mean deviation about mean = (0.2 + 2.8 + 0.8 + 2.2 + 1.2) / 5
= 7 / 5
= 1.4
Therefore, the value of mean deviation about mean for the given dataset is 1.44 (Option C).
The average age of 15 students of a class is 15 years. Out of them, the average age of 5 students is 14 years and that of the other 9 students is 16 years. The age of the 15th students is:- a)11 years
- b)14 years
- c)15 years
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The average age of 15 students of a class is 15 years. Out of them, the average age of 5 students is 14 years and that of the other 9 students is 16 years. The age of the 15th students is:
a)
11 years
b)
14 years
c)
15 years
d)
None of these
|
Nandini Patel answered |
Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.
Mean is of ———— types.- a)3
- b)4
- c)8
- d)5
Correct answer is option 'A'. Can you explain this answer?
Mean is of ———— types.
a)
3
b)
4
c)
8
d)
5
|
|
Stuti Desai answered |
Types of Mean
Mean is a statistical measure of central tendency. It is calculated by adding all the values in a given data set and dividing the sum by the total number of values. There are different types of means, which are explained below:
1. Arithmetic Mean:
Arithmetic mean is the most commonly used mean. It is calculated by adding all the values in a given data set and dividing the sum by the total number of values.
Example: Calculate the arithmetic mean of the following numbers: 2, 4, 6, 8, 10.
Solution:
The sum of the numbers is 2 + 4 + 6 + 8 + 10 = 30.
The total number of values is 5.
Arithmetic mean = sum of numbers/total number of values = 30/5 = 6.
Therefore, the arithmetic mean is 6.
2. Geometric Mean:
Geometric mean is a type of mean that is used to calculate the average of two or more numbers that are multiplied together.
Example: Calculate the geometric mean of the following numbers: 2, 4, 8.
Solution:
The product of the numbers is 2 × 4 × 8 = 64.
The total number of values is 3.
Geometric mean = nth root of the product of numbers = ∛64 = 4.
Therefore, the geometric mean is 4.
3. Harmonic Mean:
Harmonic mean is a type of mean that is used to calculate the average of two or more numbers that are reciprocals of each other.
Example: Calculate the harmonic mean of the following numbers: 2, 4, 8.
Solution:
The reciprocals of the numbers are 1/2, 1/4, and 1/8.
The sum of the reciprocals is 1/2 + 1/4 + 1/8 = 7/8.
The total number of values is 3.
Harmonic mean = total number of values/sum of reciprocals = 3/(7/8) = 24/7.
Therefore, the harmonic mean is 24/7.
4. Weighted Mean:
Weighted mean is a type of mean that is used when the values in a data set have different weights or importance.
Example: Calculate the weighted mean of the following numbers, where the weights are given in parentheses:
(2, 3), (4, 2), (6, 1), (8, 4), and (10, 2).
Solution:
The weighted sum of the numbers is (2 × 3) + (4 × 2) + (6 × 1) + (8 × 4) + (10 × 2) = 76.
The total weight is 12.
Weighted mean = weighted sum/total weight = 76/12 = 6.33.
Therefore, the weighted mean is 6.33.
Conclusion:
Therefore, the correct option is (a) 3, as there are no types of means mentioned in the question.
Mean is a statistical measure of central tendency. It is calculated by adding all the values in a given data set and dividing the sum by the total number of values. There are different types of means, which are explained below:
1. Arithmetic Mean:
Arithmetic mean is the most commonly used mean. It is calculated by adding all the values in a given data set and dividing the sum by the total number of values.
Example: Calculate the arithmetic mean of the following numbers: 2, 4, 6, 8, 10.
Solution:
The sum of the numbers is 2 + 4 + 6 + 8 + 10 = 30.
The total number of values is 5.
Arithmetic mean = sum of numbers/total number of values = 30/5 = 6.
Therefore, the arithmetic mean is 6.
2. Geometric Mean:
Geometric mean is a type of mean that is used to calculate the average of two or more numbers that are multiplied together.
Example: Calculate the geometric mean of the following numbers: 2, 4, 8.
Solution:
The product of the numbers is 2 × 4 × 8 = 64.
The total number of values is 3.
Geometric mean = nth root of the product of numbers = ∛64 = 4.
Therefore, the geometric mean is 4.
3. Harmonic Mean:
Harmonic mean is a type of mean that is used to calculate the average of two or more numbers that are reciprocals of each other.
Example: Calculate the harmonic mean of the following numbers: 2, 4, 8.
Solution:
The reciprocals of the numbers are 1/2, 1/4, and 1/8.
The sum of the reciprocals is 1/2 + 1/4 + 1/8 = 7/8.
The total number of values is 3.
Harmonic mean = total number of values/sum of reciprocals = 3/(7/8) = 24/7.
Therefore, the harmonic mean is 24/7.
4. Weighted Mean:
Weighted mean is a type of mean that is used when the values in a data set have different weights or importance.
Example: Calculate the weighted mean of the following numbers, where the weights are given in parentheses:
(2, 3), (4, 2), (6, 1), (8, 4), and (10, 2).
Solution:
The weighted sum of the numbers is (2 × 3) + (4 × 2) + (6 × 1) + (8 × 4) + (10 × 2) = 76.
The total weight is 12.
Weighted mean = weighted sum/total weight = 76/12 = 6.33.
Therefore, the weighted mean is 6.33.
Conclusion:
Therefore, the correct option is (a) 3, as there are no types of means mentioned in the question.
Each value is considered as many times as it occurs for- a)weighted average
- b)simple average
- c)both
- d)none
Correct answer is option 'A'. Can you explain this answer?
Each value is considered as many times as it occurs for
a)
weighted average
b)
simple average
c)
both
d)
none
|
|
Rajeev Chaudhary answered |
Weighted Average vs Simple Average
Weighted average and simple average are two methods of calculating the mean of a set of values. The main difference between them is that in a weighted average, each value is given a weight based on its importance or frequency, while in a simple average, all values are given equal weight.
Weighted Average
In a weighted average, each value is considered as many times as it occurs, but its contribution to the final mean is proportional to its weight. The weight can be based on any factor, such as frequency, importance, or relevance.
For example, if we have a set of test scores with the following values and frequencies:
- 80 (3 times)
- 90 (2 times)
- 70 (1 time)
To calculate the weighted average, we need to assign a weight to each value based on its frequency. In this case, the weights would be:
- 80 (weight = 3)
- 90 (weight = 2)
- 70 (weight = 1)
Then, we can calculate the weighted average using the formula:
weighted average = (value1 x weight1 + value2 x weight2 + ... + valueN x weightN) / (weight1 + weight2 + ... + weightN)
In this example, the weighted average would be:
weighted average = (80 x 3 + 90 x 2 + 70 x 1) / (3 + 2 + 1) = 81.67
Simple Average
In a simple average, all values are given equal weight, regardless of their frequency or importance. The formula for calculating the simple average is:
simple average = (value1 + value2 + ... + valueN) / N
Using the same example as before, the simple average would be:
simple average = (80 + 90 + 70) / 3 = 80
Conclusion
In summary, each value is considered as many times as it occurs in a weighted average, while in a simple average, all values are given equal weight. The choice between these two methods depends on the purpose of the analysis and the nature of the data.
Weighted average and simple average are two methods of calculating the mean of a set of values. The main difference between them is that in a weighted average, each value is given a weight based on its importance or frequency, while in a simple average, all values are given equal weight.
Weighted Average
In a weighted average, each value is considered as many times as it occurs, but its contribution to the final mean is proportional to its weight. The weight can be based on any factor, such as frequency, importance, or relevance.
For example, if we have a set of test scores with the following values and frequencies:
- 80 (3 times)
- 90 (2 times)
- 70 (1 time)
To calculate the weighted average, we need to assign a weight to each value based on its frequency. In this case, the weights would be:
- 80 (weight = 3)
- 90 (weight = 2)
- 70 (weight = 1)
Then, we can calculate the weighted average using the formula:
weighted average = (value1 x weight1 + value2 x weight2 + ... + valueN x weightN) / (weight1 + weight2 + ... + weightN)
In this example, the weighted average would be:
weighted average = (80 x 3 + 90 x 2 + 70 x 1) / (3 + 2 + 1) = 81.67
Simple Average
In a simple average, all values are given equal weight, regardless of their frequency or importance. The formula for calculating the simple average is:
simple average = (value1 + value2 + ... + valueN) / N
Using the same example as before, the simple average would be:
simple average = (80 + 90 + 70) / 3 = 80
Conclusion
In summary, each value is considered as many times as it occurs in a weighted average, while in a simple average, all values are given equal weight. The choice between these two methods depends on the purpose of the analysis and the nature of the data.
The most commonly used measure of dispersion is- a)Range
- b)Standard deviation
- c)Coefficient of variation
- d)Quartile deviation
Correct answer is option 'B'. Can you explain this answer?
The most commonly used measure of dispersion is
a)
Range
b)
Standard deviation
c)
Coefficient of variation
d)
Quartile deviation
|
|
Alok Mehta answered |
Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range. Dispersion is contrasted with location or central tendency, and together they are the most used properties of distributions.
For values lie close to the mean , the standard deviations are- a)Big
- b)Small
- c)Moderate
- d)None
Correct answer is option 'B'. Can you explain this answer?
For values lie close to the mean , the standard deviations are
a)
Big
b)
Small
c)
Moderate
d)
None
|
|
Poonam Reddy answered |
Answer :
b)
Small If the data all lies close to the mean, then the standard deviation will
be small, while if the data is spread out over a large range of values, s will be large. Having outliers will increase the standard deviation.
Each different value is considered only once for- a)simple average
- b)weighted average
- c)both
- d)none
Correct answer is option 'A'. Can you explain this answer?
Each different value is considered only once for
a)
simple average
b)
weighted average
c)
both
d)
none
|
|
Dhruba Choudhary answered |
Simple Average and Weighted Average
Simple average and weighted average are two methods of calculating the average of a set of values. Both methods are commonly used in various fields, including finance, statistics, and science. However, there are some key differences between these two methods.
Simple Average
A simple average is calculated by adding up all the values in a set and dividing the total by the number of values. For example, if you have a set of numbers 2, 4, 6, and 8, the simple average would be:
(2 + 4 + 6 + 8) / 4 = 5
In simple average, each value is considered equally important, regardless of its magnitude or significance.
Weighted Average
A weighted average, on the other hand, assigns different weights or importance to each value in a set. This means that some values have a greater impact on the average than others. Weighted averages are commonly used when some values are more important than others or when some values occur more frequently than others.
For example, suppose you have a set of grades for a course where the final exam is worth 50% of the total grade, the midterm is worth 30%, and the homework assignments are worth 20%. To calculate the weighted average of the grades, you would multiply each grade by its corresponding weight, add up the products, and divide by the total weight.
Both
In this question, it is mentioned that each different value is considered only once. This means that each value has the same importance, and there are no weights assigned to any value. Therefore, only a simple average can be used to calculate the average of the given values.
None
Option 'D' is not the correct answer as it suggests that neither simple average nor weighted average can be used to calculate the average of the given values, which is incorrect. Simple average can be used to calculate the average of any set of values, regardless of their magnitude or frequency, as long as each value is considered equally important.
Simple average and weighted average are two methods of calculating the average of a set of values. Both methods are commonly used in various fields, including finance, statistics, and science. However, there are some key differences between these two methods.
Simple Average
A simple average is calculated by adding up all the values in a set and dividing the total by the number of values. For example, if you have a set of numbers 2, 4, 6, and 8, the simple average would be:
(2 + 4 + 6 + 8) / 4 = 5
In simple average, each value is considered equally important, regardless of its magnitude or significance.
Weighted Average
A weighted average, on the other hand, assigns different weights or importance to each value in a set. This means that some values have a greater impact on the average than others. Weighted averages are commonly used when some values are more important than others or when some values occur more frequently than others.
For example, suppose you have a set of grades for a course where the final exam is worth 50% of the total grade, the midterm is worth 30%, and the homework assignments are worth 20%. To calculate the weighted average of the grades, you would multiply each grade by its corresponding weight, add up the products, and divide by the total weight.
Both
In this question, it is mentioned that each different value is considered only once. This means that each value has the same importance, and there are no weights assigned to any value. Therefore, only a simple average can be used to calculate the average of the given values.
None
Option 'D' is not the correct answer as it suggests that neither simple average nor weighted average can be used to calculate the average of the given values, which is incorrect. Simple average can be used to calculate the average of any set of values, regardless of their magnitude or frequency, as long as each value is considered equally important.
If all the values are multiplies by the same quantity, the __________ & _________ also would be multiple of the same quantity.- a)Mean, standard deviations
- b)Mean, median
- c)Mean, mode
- d)Median, deviations
Correct answer is option 'A'. Can you explain this answer?
If all the values are multiplies by the same quantity, the __________ & _________ also would be multiple of the same quantity.
a)
Mean, standard deviations
b)
Mean, median
c)
Mean, mode
d)
Median, deviations
|
|
Divya Dasgupta answered |
Ratio between the values remains the same.
For the values of a variable 3, 1, 5, 2, 6, 8, 4 the median is- a)3
- b)5
- c)4
- d)None
Correct answer is option 'C'. Can you explain this answer?
For the values of a variable 3, 1, 5, 2, 6, 8, 4 the median is
a)
3
b)
5
c)
4
d)
None
|
|
Sameer Basu answered |
Explanation:
- The median is the middle value in a set of numbers when they are arranged in order.
- To find the median of the given set of numbers, we need to arrange them in ascending order.
- Arranging the numbers in ascending order we get 1, 2, 3, 4, 5, 6, 8.
- The middle value of this set is 4.
- Therefore, the median of the given set of numbers is 4.
- Option C is the correct answer.
If the class interval is open-end then it is difficult to find- a)Frequency
- b)A.M
- c)Both
- d)None
Correct answer is option 'B'. Can you explain this answer?
If the class interval is open-end then it is difficult to find
a)
Frequency
b)
A.M
c)
Both
d)
None
|
|
Ruchi Mishra answered |
Explanation:
Open-end class interval is the interval that does not have a fixed upper or lower limit. These intervals are represented by a symbol like (<) or="" (="">). It is difficult to find the arithmetic mean (A.M) for open-end class interval because the exact values of upper and lower limits are not known. However, we can find the frequency of data even if the class interval is open-end. The following points explain why it is difficult to find the arithmetic mean for open-end class interval.
Difficulty in finding upper and lower class limits:
In open-end class interval, the exact values of upper and lower limits are not known. Therefore, it is difficult to determine the mid-value of each class interval. The mid-value is used to find the arithmetic mean of a set of data.
Difficulty in calculating the summation of mid-values:
The formula for calculating the arithmetic mean of data is the sum of all mid-values divided by the total frequency of data. In open-end class interval, it is difficult to calculate the summation of mid-values because the exact mid-value of each class interval is not known.
Difficulty in calculating the total frequency:
The total frequency of data can be calculated even if the class interval is open-end. The frequency of each class interval can be determined by counting the number of observations falling in that interval. However, the arithmetic mean cannot be calculated without the mid-values of each class interval.
Conclusion:
In conclusion, it is difficult to find the arithmetic mean for open-end class interval because the exact values of upper and lower limits are not known. However, the frequency of data can still be determined even if the class interval is open-end.
Open-end class interval is the interval that does not have a fixed upper or lower limit. These intervals are represented by a symbol like (<) or="" (="">). It is difficult to find the arithmetic mean (A.M) for open-end class interval because the exact values of upper and lower limits are not known. However, we can find the frequency of data even if the class interval is open-end. The following points explain why it is difficult to find the arithmetic mean for open-end class interval.
Difficulty in finding upper and lower class limits:
In open-end class interval, the exact values of upper and lower limits are not known. Therefore, it is difficult to determine the mid-value of each class interval. The mid-value is used to find the arithmetic mean of a set of data.
Difficulty in calculating the summation of mid-values:
The formula for calculating the arithmetic mean of data is the sum of all mid-values divided by the total frequency of data. In open-end class interval, it is difficult to calculate the summation of mid-values because the exact mid-value of each class interval is not known.
Difficulty in calculating the total frequency:
The total frequency of data can be calculated even if the class interval is open-end. The frequency of each class interval can be determined by counting the number of observations falling in that interval. However, the arithmetic mean cannot be calculated without the mid-values of each class interval.
Conclusion:
In conclusion, it is difficult to find the arithmetic mean for open-end class interval because the exact values of upper and lower limits are not known. However, the frequency of data can still be determined even if the class interval is open-end.
Chapter doubts & questions for Chapter 14: Measures of Central Tendency and Dispersion - Quantitative Aptitude for CA Foundation 2025 is part of CA Foundation exam preparation. The chapters have been prepared according to the CA Foundation exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for CA Foundation 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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