Class 11 RD Sharma Solutions- Chapter 32 Statistics - Exercise 32.6

Question 1. Calculate the mean and S.D. for the following data:

Solution:

Given: 

Number of observations, N = 90 and A = 25

\sum f_iu_i =10\\ \sum f_iu_i^2  =150    

h = 10

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i) 

\overline{x} = 25 + 10(10/90) = 26.11 

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 10[(150/90) - (10/90)²]

= 165.4 

Standard Deviation = √var(x) = √165.4 = 12.86  

Question 2. Calculate the standard deviation for the following data:

Solution:

Given: N = 300 and A =105

\sum f_iu_i =137\\ \sum f_iu_i^2  =665   

h = 30

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i) 

\overline{x} = 105 + 30(137/300) = 118.7

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2] 

= 900[(665/300) - (137/300)²] 

= 1807.31 

Standard Deviation = √var(x) = √1807.31 = 42.51

Question 3. Calculate the A.M. and S.D. for the following distribution:

Solution:

Given:  N = 79 and A =35

\sum f_iu_i =-71\\ \sum f_iu_i^2  =305   

h = 10

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i) 

\overline{x} = 35 + 10(-71/79) = 26.01

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 100[(305/79) - (-71/79)²] 

= 305.30

Standard Deviation  = √var(x) = √305.30 = 17.47

Question 4. A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure is 40. Find the correct mean and S.D.

Solution:

According to question, we have, 

n = 100 , \overline{x} = 40 , \sigma = 5.1 \\ \overline{x} = \frac{1}{n}\sum x_i \\ \sum x_i = n\overline{x} = 4000 \\ Incorrect \sum x_i = 4000

And, also 

\sigma = 5.1 \\ \sigma^2 = 26.01 \\ \frac{1}{n}\sum x_i^2 - Mean^2 = 26.01 \\ \frac{1}{100}\sum x_i^2 - 1600 = 26.01 \\

\sum x_i^2 = 1626.01 x 100 

Incorrected \sum x_i^2 = 162601

On replacing the incorrect observation of 50 by 40, we get,

Incorrect \sum x_i  = 4000 

Corrected \sum x_i = 4000 - 50 + 40 = 3990

Incorrected \sum x_i^2 = 162601 

Corrected \sum x_i^2 = 162601 - 50² + 40² = 161701

Now, we have,

Corrected Mean = 39.90

Corrected Variance = (1/100)(Corrected \sum x_i^2) - (Corrected mean)² 

 = \frac{161701}{100} - (\frac{3990}{100})^2 \\ = \frac{161701 * 100 - (3990)^2}{100^2} \\ = \frac{16170100-15920100}{10000}

= 25

Corrected standard deviation = √25 = 5

Question 5. Calculate the mean, median, and standard deviation of the following distribution:

Solution:

Now, using the given values, we have

Mean = 53 + 5 x (-30/50) 

= 50

Variance = 25 x ((134/50) - (9/25)

= 58

Standard Deviation = √58 

= 7.62

Question 6. Find the mean and variance of frequency distribution given below:

Solution:

The data can be converted to a continuous frequency distribution by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each of the class interval. 

Class Interval	fi	xi	ui	fiui	ui2	fiui2
1 - 2	6	1.5	-4	-24	16	96
3 - 4	4	3.5	-2	-8	4	16
5 - 6	5	5.5	0	0	0	0
7 - 8	1	7.5	2	2	4	4
	N = 16			Total = -30		Total = 116

Given: N = 16 and A = 5.5

\sum f_iu_i =-30\\ \sum f_iu_i^2  =116   and h=1

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i) 

\overline{x} = 5.5 + 1((1/6) x (-30))

= 3.625 

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2] 

= 1 [((1/16) x 116) - ((1/16) x (-30)²] 

= 3.74

Question 7. The weight of coffee in 70 jars is shown in the following table :

Calculate mean, variance, and standard deviation.

Solution:

Now, using the given values, we have

Mean = 203.5 + 2 x (-54/70)

= 201.9

Variance = 4 x (62/70) - (-54/70)

= 0.98

Standard Deviation = √0.98

= 0.099

Question 8. Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Solution:

Mean = 40

Standard Deviation = 10

n = 100

\sum x_i = 40 * 100 = 4000

Corrected Sum = 4000 – 30 +70 + 3 + 27 = 3930

Corrected mean = 39.3

Variance = 100

100 = \frac{\sum x_i^2}{100} -(40)^2 

Incorrect \sum x_i^2 = 170000 

So, Corrected \sum x_i^2 = Incorrect \sum x_i^2 - (Sum of squares of incorrect values) + 

                                                                                    (Sum of squares of corrected values) 

Corrected\sum x_i^2 = 170000 - (900 + 4900) + (9+729)  

= 164938

Corrected \space \sigma = \sqrt{\frac{Corrected \space \sum x_i^2}{n}-(Corrected \space Mean)^2} \\ = \sqrt{\frac{164938}{100}-(39.3)^2} \\

 = 10.24

Question 9. While calculating the mean and variance of 10 reading, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and variance. 

Solution:

Mean = 45

Variance = 16

n = 10

\sum x_i = 450

So, Corrected Sum = 450 – 52 + 25 = 423

Corrected mean = 42.3

Variance = 16

116 = \frac{\sum x_i^2}{10} -(45)^2 \\ Incorrect \sum x_i^2 = 20410

Corrected \sum x_i^2 = Incorrect  \sum x_i^2- (Sum of squares of incorrect values) + 

                                                                             (Sum of squares of corrected values) 

Corrected \sum x_i^2  = 20410 - 2704 + 625 = 18331

Corrected \space \sigma = \sqrt{\frac{Corrected \space \sum x_i^2}{n}-(Corrected \space Mean)^2} \\ = \sqrt{\frac{18331}{10}-(42.3)^2}

= 6.62

So, Corrected variance = 6.62 * 6.62 = 43.82

Question 10. Calculate mean, variance, and standard deviation of the following frequency distribution:

Solution:

Given:

Number of observations, N = 70 and A = 35

\sum f_iu_i =-98\\ \sum f_iu_i^2  =250

h = 10

Mean = \overline{x} = A+h(\frac{1}{N}\sum f_iu_i) 

\overline{x} = 35 + 10(-98/70) = -21

var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]

= 100[(1/70) x 250 - (1/70) x (-98)²]

= 161 

Standard Deviation = √var(x) 

= √161

= 12.7

Summary

Exercise 32.6 of RD Sharma Class 11 Statistics chapter deals with calculating measures of central tendency (mean) and dispersion (variance, standard deviation) for different types of data. Students are required to apply formulas and methods to find these values.