Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 13. Explain,  by taking a suitable example, how the arithmetic mean alters by

(i) adding a constant k to each term,
(ii) Subtracting a constant k from each term,
(iii) multiplying each term by a constant k and
(iv) dividing each term by non-zero constant k.

SOLUTION :

Let say numbers are 3 , 4 , 5

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(i). Adding constant term k = 2 in each term.

New numbers are = 5 , 6 , 7

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

∴ new  mean will be 2 more than the original mean.

(ii). Subtracting constant term k = 2 in each term.

New numbers are = 1 , 2 , 3

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

∴ new  mean will be 2 less than the original mean.

(iii) . Multiplying by constant term k = 2 in each term.

New numbers are = 6 , 8 , 10

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

∴ new  mean will be 2 times of  the original mean.

(iv) . Divide the constant term k =2 in each term.

New numbers are = 1.5 , 2 , 2.5.

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

∴ new  mean will be half  of  the original mean.


Q 14. The mean of marks scored by 100 students was found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct mean.

SOLUTION :

Mean marks of 100 students = 40

Sum of marks of 100 students = 100×40

= 4000

Correct value = 53

Incorrect value = 83

Correct sum = 4000 – 83 + 53 = 3970

∴ correct  mean = 3970/100 = 39.7


Q 15 . The traffic police recorded the speed (in km/hr) of 10 motorists as 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 . Later on, an error in recording instrument was found. Find the correct average speed of the motorists if the instrument is recorded 5 km/hr less in each case.

SOLUTION :

The speed of 10 motorists are 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 .

Later on it was discovered that the instrument recorded 5 km/hr less than in each case

∴ correct values are = 52 , 58 , 54 , 65 , 44 , 47 , 60 , 62 , 57 , 53.

∴ correct  mean  RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics


Q 16. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

SOLUTION :

The mean of five numbers is 27

The sum of five numbers = 5×27 = 135

If one number is excluded , the new mean is 25

∴Sum of 4 numbers = 4×25 = 100

∴ Excluded number = 135 – 100 = 35


Q 17. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

SOLUTION :

The mean weight per student in a group of 7 students = 55 kg

Weight of 6 students (in kg) =  52 , 54 , 55 , 53 , 56 and 54

Let the weight of seventh student = x kg

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

∴ weight of seventh student = 61 kg.


Q 18. The mean weight of 8 numbers is 15. If each number is multiplied by 2 , what will be the new mean?

SOLUTION :

We have ,

The mean weight of 8 numbers is 15

Then , the sum of 8 numbers = 8×15 = 120

If each number is multiplied by 2

Then , new mean = 120×2 = 240

∴ new  mean = 240/8 = 30.


Q 19. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

SOLUTION :

The mean of 5 numbers is 18

Then , the sum of 5 numbers = 5×18 = 90

If one number is excluded

Then , the mean of 4 numbers = 16

∴ sum of 4 numbers = 4×16 = 64

Excluded number = 90 – 64 = 26.


Q 20. The mean of 200 items was 50. Later on, it was on discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

SOLUTION :

The mean of 200 items = 50

Then the sum of 200 items = 200×50 = 10,000

Correct values = 192 and 88.

Incorrect values = 92 and 8.

∴ correct sum = 10000 – 92 – 8 + 192 + 88 = 10180

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics


Q 21 . Find the values of n and  RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics in each of the following cases :

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

SOLUTION :

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ (x1−12)+(x2−12)+ ……… +(xn−12) = −10

⇒(x1+x2+x3+x4+x5+⋅⋅⋅+xn)−(12+12+12+12+⋅⋅⋅⋅+12)=−10

⇒ ∑ x−12n =−10⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ (x1−3)+(x2−3)+⋅⋅⋅⋅⋅+(xn−3) = 62

⇒ (x1+x2+⋅⋅⋅+xn)−(3+3+3+⋅⋅⋅+3) = 62

⇒ ∑x−3n = 62⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

By subtracting equation (1) from equation(2) , we get

∑x−3n−∑x+12n  62+10

⇒ 9n = 72

⇒ n = 72/9 = 8

Put value of n in equation (1)

∑x−12×8 = −10

⇒∑x−96 = −10

⇒∑x = 96−10 = 86

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(x1−10)+(x2−10)+ ………… + (xn−10) = 30

⇒(x1+x2+x3+x4+x5+⋅⋅⋅+xn)−(10+10+10+10+⋅⋅⋅⋅+10) = 30

⇒ ∑x−10n = 30⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒(x1−6)+(x2−6)+⋅⋅⋅⋅⋅+(xn−6) = 150

⇒(x1+x2+⋅⋅⋅+xn)−(6+6+6+⋅⋅⋅+6) = 150

⇒∑x−6n = 150⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

By subtracting equation (1) from equation(2) , we get

∑x−6n−∑x+10n = 150−30

⇒ 4n = 120

⇒ n = 120/4 = 30

Put value of n in equation (1)

∑x−10×30 = 30

⇒ ∑x−300 = 30

⇒ ∑x=30+300 = 330

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics


Q 22 . The sum of the deviations of a set of n values x1,x2,x3,⋅⋅⋅,xn measured from 15 and -3 are -90 and 54 respectively . Find the value of n and mean .

SOLUTION :

Given :

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ (x1−15)+(x2−15)+⋅⋅⋅⋅⋅+(xn−15) = −90

⇒ (x1+x2+⋅⋅⋅⋅+n)−(15+15+15+⋅⋅⋅⋅⋅⋅+15) =−90

⇒ ∑x−15n =−90⋅⋅⋅⋅⋅(1)

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

⇒ (x1+3)+(x2+3)+⋅⋅⋅⋅⋅+(xn+3) = 54

⇒ (x1+x2+⋅⋅⋅⋅+n)+(3+3+3+⋅⋅⋅⋅⋅⋅+3) = 54

⇒ ∑x+3n = 54⋅⋅⋅⋅⋅(2)

By subtracting equation (1) from equation(2) , we get

∑x+3n−∑x+15n = 54+90

⇒ 18n =144

⇒ n =144/18 = 8

Put value of n in equation(1)

∑x−15×8 = −90

∑x−120 = −90

∑x = 120−90 = 30

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics


Q 23 . Find the sum of the deviations of the variate values 3 , 4 , 6 , 7 , 8 , 14 from their mean.

SOLUTION :

Values 3 , 4 , 6 , 7 , 8 , 14

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

= 7 

∴ Sum of deviation of values from their mean

= (3−7)+(4−7)+(6−7)+(7−7)+(8−7)+(14−7)

= – 4 – 3 – 1 + 0 + 1 + 7

= – 8 + 8 = 0


Q 24 . If RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is the mean of the ten natural numbers x1,x2,x3,⋅⋅⋅,x10 show that 

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

SOLUTION :

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics    [Byequation(i)]

RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

The document RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-24.1, (Part - 2), Measures Of Central Tendency, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are the measures of central tendency in statistics?
Ans. Measures of central tendency are statistical measures that represent the center or average of a data set. The three commonly used measures of central tendency are the mean, median, and mode. The mean is the arithmetic average, calculated by summing all the values in the data set and dividing by the total number of values. The median is the middle value when the data set is arranged in ascending or descending order. The mode is the value or values that appear most frequently in the data set.
2. How is the mean calculated?
Ans. The mean is calculated by summing all the values in the data set and dividing by the total number of values. For example, if we have the data set {2, 4, 6, 8, 10}, the mean is calculated as (2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6. Therefore, the mean of this data set is 6.
3. What is the median and how is it calculated?
Ans. The median is the middle value when the data set is arranged in ascending or descending order. To calculate the median, we first arrange the data set in order. If the total number of values is odd, the median is the middle value. If the total number of values is even, the median is the average of the two middle values. For example, if we have the data set {3, 6, 7, 9, 12}, the median is 7, as it is the middle value. If we have the data set {2, 4, 6, 8, 10, 12}, the median is (6 + 8) / 2 = 7, as it is the average of the two middle values.
4. How is the mode determined in statistics?
Ans. The mode is determined by identifying the value or values that appear most frequently in the data set. In other words, it is the value that has the highest frequency. For example, if we have the data set {2, 3, 4, 4, 6, 6, 6, 8}, the mode is 6, as it appears three times, which is more than any other value in the data set.
5. Can a data set have more than one mode?
Ans. Yes, a data set can have more than one mode. If there are two or more values that appear with the same highest frequency, then the data set is said to be multimodal. For example, if we have the data set {2, 2, 4, 4, 6, 6, 8, 8}, both 2 and 4 are modes as they appear twice, which is more than any other value in the data set.
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