Class 9 Exam > Class 9 Notes > RD Sharma Solutions for Class 9 Mathematics > RD Sharma Solutions Ex-24.1, (Part - 1), Measures Of Central Tendency, Class 9, Maths
RD Sharma Solutions Ex-24.1, (Part - 1), Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download
Q.1: If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively. Find the mean height.
SOLUTION :
Given : the heights of 5 persons are 140 cm , 150 cm , 152 cm , 158 cm and 161 cm
Q 2 . Find the mean of 994 , 996 , 998 , 1000 , 1002.
SOLUTION :
Given :
Numbers are 994 , 996 , 998 , 1000 , 1002.
Q 3 . Find the mean of first five natural numbers.
SOLUTION :
The first five odd numbers are 1 , 2 , 3 , 4 , 5.
Q 4 . Find the mean of all factors of 10.
SOLUTION :
All factors of 6 are 1 , 2 , 5 , 10.
Mean = 4.5
Q 5 . Find the mean of first ten even natural numbers.
SOLUTION :
The first five even natural numbers are 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20
Mean = 11
Q 6 . Find the mean of x , x + 2 , x + 4 , x + 6 , x + 8.
SOLUTION :
Numbers are x , x + 2 , x + 4 , x + 6 , x + 8.
Q 7. Find the mean of first five multiples of 3.
SOLUTION :
First five multiples of 3 are 3 , 6 , 9 , 12 , 15.
= 9
Mean = 9
Q 8 . Following are the weights of 10 new born babies in a hospital on a particular day : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6 (in kg). Find the mean.
SOLUTION : The weights (in kg) of 10 new born babies are : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6
= 4 kg
Q 9 . The percentage marks obtained by students of a class in mathematics are as follows: 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1 .Find their mean.
SOLUTION :
The percentage marks obtained by students are 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1
Mean Marks = 39.5
Q 10. The numbers of children in 10 families of a locality are 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5 . Find the number of children per family.
SOLUTION :
The numbers of children in 10 families are : 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5
Q 11 . If M is the mean of x1,x2,x3,x4,x5 and x6 , Prove that
(x1−M)+(x2−M)+(x3−M)+(x4−M)+(x5−M)+(x6−M) = 0.
SOLUTION :
Let M be the mean of x1,x2,x3,x4,x5 and x6
= x1+x2+x3+x4+x5+x6 = 6M
To Prove :- (x1−M)+(x2−M)+(x3−M)+(x4−M)+(x5−M)+(x6−M) = 0.
Proof :- L . H . S
= (x1−M)+(x2−M)+(x3−M)+(x4−M)+(x5−M)+(x6−M)
= (x1+x2+x3+x4+x5+x6)−(M+M+M+M+M+M)
= 6M – 6M
= 0
= R . H . S
Q 12 . Duration of sunshine(in hours) in Amritsar for first 10 days of August 1997 as reported by the Meterological Department are given as follows : 9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9
1. Find the mean 
2.Verify that 
SOLUTION :
Duration of sunshine (in hours ) for 10 days are =9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9
= 56/10 = 5.6
= (9.6−5.6)+(5.2−5.6)+(3.5−5.6)+(1.5−5.6)+(1.6−5.6)+(2.4−5.6)+(2.6−5.6)+(8.4−5.6)+(10.3−5.6)+(10.9−5.6)
= 4 – 0.4 – 2.1 – 4.1 – 4 – 3.2 – 3 + 2.8 + 4.7 + 5.3
= 16.8 – 16.8 = 0
= R.H.S
The document RD Sharma Solutions Ex-24.1, (Part - 1), 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 - 1), Measures Of Central Tendency, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics
| 1. What are measures of central tendency in statistics? |  |
Ans. Measures of central tendency are statistical measures that represent the center or typical value of a dataset. These measures include the mean, median, and mode. The mean is the average value of the dataset, the median is the middle value when the data is arranged in ascending or descending order, and the mode is the value that appears most frequently in the dataset.
| 2. How is the mean calculated for a given dataset? | |
Ans. The mean is calculated by summing up all the values in the dataset and dividing the sum by the total number of values. Mathematically, the formula for calculating the mean is:
Mean = (sum of all values) / (total number of values)
For example, if we have a dataset {5, 7, 9, 11, 13}, the mean would be calculated as (5+7+9+11+13)/5 = 9.
| 3. What is the median and how is it determined in a dataset? | |
Ans. The median is the middle value in a dataset when the data is arranged in ascending or descending order. To determine the median, the data is first arranged in order, and then the middle value is identified. If the dataset has an odd number of values, the middle value is the median. If the dataset has an even number of values, the median is the average of the two middle values.
For example, in the dataset {5, 7, 9, 11, 13}, the median would be 9. In the dataset {5, 7, 9, 11, 13, 15}, the median would be (9+11)/2 = 10.
| 4. How is the mode determined in a dataset? | |
Ans. The mode in a dataset is the value that appears most frequently. To determine the mode, we count the frequency of each value in the dataset and identify the value with the highest frequency. It is possible for a dataset to have multiple modes if multiple values have the same highest frequency.
For example, in the dataset {5, 7, 9, 9, 11, 13, 13, 13}, the mode would be 13 as it appears three times, which is the highest frequency.
| 5. How do measures of central tendency help in understanding a dataset? | |
Ans. Measures of central tendency provide valuable information about the typical or central value of a dataset. The mean, median, and mode help in understanding the average, middle, and most frequently occurring values, respectively. These measures help in summarizing the data, making comparisons, and drawing conclusions. They are also useful in identifying outliers or extreme values that may impact the overall interpretation of the dataset.
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