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

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

Expenditure (in ₹):	0-10	10-20	20-30	30-40	40-50
Frequency:	14	13	27	21	15

Solution:

CI	f	x	u = (x – A)/h	fu	u2	fu2
0 – 10	14	5	-2	-28	4	56
10 – 20	13	15	-1	-13	1	13
20 – 30	27	25	0	0	0	0
30 – 40	21	35	1	21	1	21
40 – 50	15	45	2	30	4	60
	90			10		150

Given:

Number of observations, N = 90 and A = 25

h = 10

Mean = 

 = 25 + 10(10/90) = 26.11 

= 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:

Class:	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequency:	9	17	43	82	81	44	24

Solution:

CI	f	x	u = (x – A)/h	f × u	u2	fu2
0 – 30	9	15	-3	-27	9	81
30 – 60	17	45	-2	-34	4	68
60 – 90	43	75	-1	-43	1	43
90 – 120	82	105	0	0	0	0
120 – 150	81	135	1	81	1	81
150 – 180	44	165	2	88	4	176
180 – 210	24	195	3	72	9	216
	300			137		665

Given: N = 300 and A =105

h = 30

Mean = 

 = 105 + 30(137/300) = 118.7

= 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:

Class:	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency:	18	16	15	12	10	5	2	1

Solution:

CI	f	x	u = (x – A)/h	f × u	u2	fu2
0 – 10	18	5	-3	-54	9	162
10 – 20	16	15	-2	-32	4	64
20 – 30	15	25	-1	-15	1	15
30 – 40	12	35	0	0	0	0
40 – 50	10	45	1	10	1	10
50 – 60	5	55	2	10	4	20
60 – 70	2	65	3	6	9	18
70 – 80	1	75	4	4	16	16
	79			-71		305

Given:  N = 79 and A =35

h = 10

Mean = 

 = 35 + 10(-71/79) = 26.01

= 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 , 

And, also 

 = 1626.01 x 100 

Incorrected  = 162601

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

Incorrect  = 4000 

Corrected  = 4000 – 50 + 40 = 3990

Incorrected = 162601 

Corrected  = 162601 – 50² + 40² = 161701

Now, we have,

Corrected Mean = 39.90

Corrected Variance = (1/100)(Corrected ) – (Corrected mean)²

= 25

Corrected standard deviation = √25 = 5

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

Class-interval	31-35	36-40	41-45	46-50	51-55	56-60	61-65	66-70
Frequency:	2	3	8	12	16	5	2	3

Solution:

CI	Freq	Mid Value	ui	fiui	fiui2
31 – 35	2	33	-4	-8	32
36 – 40	3	38	-3	-9	27
41 – 45	8	43	-2	-16	32
46 – 50	12	48	-1	-12	12
51 – 55	16	53	0	0	0
56 – 60	5	58	1	5	5
61 – 65	2	63	2	4	8
66 – 70	2	68	3	6	18
	N = 50			Total = – 30	Total = 134

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:

xi	1 ≤ x < 3	3 ≤ x < 5	5 ≤ x < 7	7 ≤ x < 9
fi	6	4	5	1

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

 and h=1

Mean = 

 = 5.5 + 1((1/6) x (-30))

= 3.625 

= 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 :

Weight (in grams)	200-201	201-202	202-203	203-204	204-205	205-206
Frequency	13	27	18	10	1	1

Calculate mean, variance, and standard deviation.

Solution:

CI	xi	fi	ui	fiui	fiui2
200 – 201	200.5	13	-15	-19.5	29.25
201 – 202	201.5	27	-1	-27	27
202 – 203	202.5	18	-0.5	-9	4.5
203 – 204	203.5	10	0	0	0
204 – 205	204.5	1	0.5	0.5	0.25
205 – 206	205.5	1	1	1	1
		N = 70		Total = – 54	Total = 62

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

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

Corrected mean = 39.3

Variance = 100

Incorrect \sum x_i^2 = 170000 

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

                                                                                    (Sum of squares of corrected values) 

Corrected = 170000 – (900 + 4900) + (9+729)  

= 164938

 = 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

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

Corrected mean = 42.3

Variance = 16

1

Corrected = Incorrect  – (Sum of squares of incorrect values) + 

                                                                             (Sum of squares of corrected values) 

Corrected   = 20410 – 2704 + 625 = 18331

= 6.62

So, Corrected variance = 6.62 * 6.62 = 43.82

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

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	11	29	18	4	5	3

Solution:

CI	xi	fi	ui	fiui	fiui2
0-10	5	11	-3	-33	99
10-20	15	29	-2	-58	116
20-30	25	18	-1	-18	18
30-40	35	4	0	0	0
40-50	45	5	1	5	5
50-60	55	3	2	6	12
		N = 70		Total = – 98	Total = 250

Given:

Number of observations, N = 70 and A = 35

h = 10

Mean = 

 = 35 + 10(-98/70) = -21

= 100[(1/70) x 250 – (1/70) x (-98)²]

= 161 

Standard Deviation = √var(x) 

= √161

= 12.7