NCERT Solutions for Class 11 Maths Chapter 13 Statistics

Exercise 1 : 4, 7, 8, 9, 10, 12, 13, 17

First, we have to find (x̅) of the given data.

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 10 – 4 = 6, 10 – 7 = 3, 10 – 8 = 2, 10 – 9 = 1, 10 – 10 = 0,

10 – 12 = – 2, 10 – 13 = – 3, 10 – 17 = – 7

6, 3, 2, 1, 0, -2, -3, -7

Now, absolute values of the deviations,

6, 3, 2, 1, 0, 2, 3, 7

MD = sum of deviations/ number of observations

= 24/8

= 3

So, the mean deviation for the given data is 3.

Exercise 2 : 2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

First, we have to find (x̅) of the given data.

So, the respective values of the deviations from mean,

i.e., xi – x̅ are, 50 – 38 = -12, 50 -70 = -20, 50 – 48 = 2, 50 – 40 = 10, 50 – 42 = 8,

50 – 55 = – 5, 50 – 63 = – 13, 50 – 46 = 4, 50 – 54 = -4, 50 – 44 = 6

-12, 20, -2, -10, -8, 5, 13, -4, 4, -6

Now, absolute values of the deviations,

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

MD = sum of deviations/ number of observations

= 84/10

= 8.4

So, the mean deviation for the given data is 8.4.

 

First, we have to arrange the given observations into ascending order.

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

The number of observations is 12.

Then,

Median = ((12/2)th observation + ((12/2)+ 1)th observation)/2

(12/2)th observation = 6th = 13

(12/2)+ 1)th observation = 6 + 1

= 7th = 14

Median = (13 + 14)/2

= 27/2

= 13.5

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

Mean Deviation

= (1/12) × 28

= 2.33

So, the mean deviation about the median for the given data is 2.33.

 

First, we have to arrange the given observations into ascending order.

36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

The number of observations is 10.

Then,

Median = ((10/2)th observation + ((10/2)+ 1)th observation)/2

(10/2)th observation = 5th = 46

(10/2)+ 1)th observation = 5 + 1

= 6th = 49

Median = (46 + 49)/2

= 95

= 47.5

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Mean Deviation

= (1/10) × 70

= 7

So, the mean deviation about the median for the given data is 7.

 

Let us make the table of the given data and append other columns after calculations.

The sum of calculated data,

The absolute values of the deviations from the mean, i.e., |xi – x̅|, as shown in the table.

 

Let us make the table of the given data and append other columns after calculations.

 

7.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Now, N = 26, which is even.

Median is the mean of the 13th and 14th observations. Both of these observations lie in the cumulative frequency of 14, for which the corresponding observation is 7.

Then,

Median = (13th observation + 14th observation)/2

= (7 + 7)/2

= 14/2

= 7

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

8.

Solution:-

Let us make the table of the given data and append other columns after calculations.

Now, N = 29, which is odd.

So, 29/2 = 14.5

The cumulative frequency greater than 14.5 is 21, for which the corresponding observation is 30.

Then,

Median = (15th observation + 16th observation)/2

= (30 + 30)/2

= 60/2

= 30

So, the absolute values of the respective deviations from the median, i.e., |xi – M| are shown in the table.

 

First, we have to write the first 10 multiples of 3.

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

So, x̅ = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

Let us make the table of the data and append other columns after calculations.

Then, the Variance

= (1/10) × 742.5

= 74.25

∴Mean = 16.5 and Variance = 74.25

 

9.

Solution:-

Let us make the table of the given data and append other columns after calculations.

10.

Solution:-

Let us make the table of the given data and append other columns after calculations.

 

Let us make the table of the given data and append other columns after calculations.

The class interval containing Nth/2 or 25th item is 20-30.

So, 20-30 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 20, c = 14, f = 14, h = 10 and n = 50

Median = 20 + (((25 – 14))/14) × 10

= 20 + 7.85

= 27.85

 

The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.

The class interval containing Nth/2 or 50th item is 35.5 – 40.5

So, 35.5 – 40.5 is the median class.

Then,

Median = l + (((N/2) – c)/f) × h

Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100

Median = 35.5 + (((50 – 37))/26) × 5

= 35.5 + 2.5

= 38

 

 

So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

Let us make the table of the given data and append other columns after calculations.

We know that the Variance,

σ2 = (1/8) × 74

= 9.2

∴Mean = 9 and Variance = 9.25

 

We know that Mean = Sum of all observations/Number of observations

∴Mean, x̅ = ((n(n + 1))2)/n

= (n + 1)/2

and also, WKT Variance,

By substituting the value of x̅, we get

WKT, (a + b)(a – b) = a2 – b2

σ2 = (n2 – 1)/12

∴Mean = (n + 1)/2 and Variance = (n2 – 1)/12

 

Let us make the table of the given data and append other columns after calculations.

 

Let us make the table of the given data and append other columns after calculations.

 

Let the assumed mean A = 64. Here, h = 1

We obtain the following table from the given data.

Mean,

Where A = 64, h = 1

So, x̅ = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, the variance,

σ2 = (12/1002) [100(286) – 02]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

∴ Mean = 64 and Standard Deviation = 1.691

 

7.

Solution:-

Let us make the table of the given data and append other columns after calculations.

8.

Solution:-

Let us make the table of the given data and append other columns after calculations.

 

Let the assumed mean, A = 92.5 and h = 5

Let us make the table of the given data and append other columns after calculations.

Mean,

Where, A = 92.5, h = 5

So, x̅ = 92.5 + ((6/60) × 5)

= 92.5 + ½

= 92.5 + 0.5

= 93

Then, the Variance,

σ2 = (52/602) [60(254) – 62]

= (1/144) [15240 – 36]

= 15204/144

= 1267/12

= 105.583

Hence, standard deviation = σ = √105.583

= 10.275

∴ Mean = 93, variance = 105.583 and Standard Deviation = 10.275

 

Let the assumed mean, A = 42.5 and h = 4

Let us make the table of the given data and append other columns after calculations.

Mean,

Where, A = 42.5, h = 4

So, x̅ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, the Variance,

σ2 = (42/1002)[100(199) – 252]

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553

∴ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.

 

For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

Co-efficient of variation (C.V.) = (σ/ x̅) × 100

Where, σ = standard deviation, x̅ = mean

For Group A.

Where A = 45,

and yi = (xi – A)/h

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ2 = (102/1502) [150(342) – (-6)2]

= (100/22500) [51,300 – 36]

= (100/22500) × 51264

= 227.84

Hence, standard deviation = σ = √227.84

= 15.09

∴ C.V for group A = (σ/ x̅) × 100

= (15.09/44.6) × 100

= 33.83

Now, for group B.

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

σ2 = (102/1502) [150(366) – (-6)2]

= (100/22500) [54,900 – 36]

= (100/22500) × 54,864

= 243.84

Hence, standard deviation = σ = √243.84

= 15.61

∴ C.V for group B = (σ/ x̅) × 100

= (15.61/44.6) × 100

= 35

By comparing the C.V. of group A and group B.

C.V of Group B > C.V. of Group A

So, Group B is more variable.

 

From the given data,

Let us make the table of the given data and append other columns after calculations.

We have to calculate Mean for x,

Mean x̅ = ∑xi/n

Where, n = number of terms

= 510/10

= 51

Then, Variance for x =

= (1/102)[(10 × 26360) – 5102]

= (1/100) (263600 – 260100)

= 3500/100

= 35

WKT Standard deviation = √variance

= √35

= 5.91

So, co-efficient of variation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have to calculate Mean for y,

Mean ȳ = ∑yi/n

Where, n = number of terms

= 1050/10

= 105

Then, Variance for y =

= (1/102)[(10 × 110290) – 10502]

= (1/100) (1102900 – 1102500)

= 400/100

= 4

WKT Standard deviation = √variance

= √4

= 2

So, co-efficient of variation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of X and Y,

C.V of X > C.V. of Y

So, Y is more stable than X.

 

(i) From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 × 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 × 5253

= Rs 34,03,944

So, firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that, standard deviation (σ)= √100

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=√(121 )

=11

Hence, the standard deviation is more in case of Firm B. That means, in firm B, there is greater variability in individual wages.

 

From the given data,

Let us make the table of the given data and append other columns after calculations.

C.V. of firm B is greater.

∴ Team A is more consistent.

 

First, we have to calculate Mean for Length x.

 

 

From the question, it is given that, n observations are x1, x2,…..xn

Mean of the n observation = x̅

Variance of the n observation = σ2

As we know,

 

(i) If the wrong item is omitted,

From the question, it is given that

The number of observations, i.e., n = 20

The incorrect mean = 20

The incorrect standard deviation = 2

 

(ii) If it is replaced by 12,

From the question, it is given that

The number of incorrect sum observations, i.e., n = 200

The correct sum of observations n = 200 – 8 + 12

n = 204

Then, correct mean = correct sum/20

= 204/20

= 10.2

 

From the question, it is given that

Mean of Mathematics = 42

Standard deviation of Mathematics = 12

Mean of Physics = 32

Standard deviation of physics = 15

Mean of Chemistry = 40.9

Standard deviation of chemistry = 20

As we know,

 

From the question, it is given that

The total number of observations (n) = 100

Incorrect mean, (x̅) = 20

And, Incorrect standard deviation (σ) = 3