NCERT Solutions for Class 11 Maths Chapter 15 Statistics

The NCERT Solutions For Class 11 Maths Chapter 15 Statistics are available as a PDF on this page. The NCERT Solutions are authored by the most experienced educators in the teaching industry, according to the latest CBSE Syllabus 2022-23 making the solution of every problem straightforward and justifiable. Every solution written in the PDF given underneath empowers the students to get ready for the exam and accomplish it. These solutions assist a Class 11 student in mastering the concepts of statistics that are categorized under the of Class 11 Maths.

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Get detailed solutions for all the questions listed under the below exercises:

Exercise 15.1 Solutions : 12 Questions

Exercise 15.2 Solutions : 10 Questions

Exercise 15.3 Solutions : 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

Access Solutions for Class 11 Maths Chapter 15 Statistics

Exercise 15.1 Page: 360

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Solution:-

First we have to find (xÌ…) of the given data

So, the respective values of the deviations from mean,

i.e., xi â€“ xÌ… are, 10 â€“ 4 = 6, 10 â€“ 7 = 3, 10 â€“ 8 = 2, 10 â€“ 9 = 1, 10 â€“ 10 = 0,

10 â€“ 12 = â€“ 2, 10 â€“ 13 = â€“ 3, 10 â€“ 17 = â€“ 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.

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

Solution:-

First we have to find (xÌ…) of the given data

So, the respective values of the deviations from mean,

i.e., xi â€“ xÌ… are, 50 â€“ 38 = -12, 50 -70 = -20, 50 â€“ 48 = 2, 50 â€“ 40 = 10, 50 â€“ 42 = 8,

50 â€“ 55 = â€“ 5, 50 â€“ 63 = â€“ 13, 50 â€“ 46 = 4, 50 â€“ 54 = -4, 50 â€“ 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.

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:-

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 â€“ 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) Ã— 28

= 2.33

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

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:-

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 â€“ 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) Ã— 70

= 7

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

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

 xi 5 10 15 20 25 fi 7 4 6 3 5

Solution:-

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

 Xi fi fixi |xi â€“ xÌ…| fi |xi â€“ xÌ…| 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 25 350 158

The sum of calculated data,

The absolute values of the deviations from the mean, i.e., |xi â€“ xÌ…|, as shown in the table.

6.

 xi 10 30 50 70 90 fi 4 24 28 16 8

Solution:-

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

 Xi fi fixi |xi â€“ xÌ…| fi |xi â€“ xÌ…| 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 80 4000 1280

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

 xi 5 7 9 10 12 15 fi 8 6 2 2 2 6

Solution:-

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

 Xi fi c.f. |xi â€“ M| fi |xi â€“ M| 5 8 8 2 16 7 6 14 0 0 9 2 16 2 4 10 2 18 3 6 12 2 20 5 10 15 6 26 8 48

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 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 â€“ M| are shown in the table.

8.

 xi 15 21 27 30 35 fi 3 5 6 7 8

Solution:-

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

 Xi fi c.f. |xi â€“ M| fi |xi â€“ M| 15 3 3 15 45 21 5 8 9 45 27 6 14 3 18 30 7 21 0 0 35 8 29 5 40

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 â€“ M| are shown in the table.

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

 Income per day in â‚¹ 0 â€“ 100 100 â€“ 200 200 â€“ 300 300 â€“ 400 400 â€“ 500 500 â€“ 600 600 â€“ 700 700 â€“ 800 Number of persons 4 8 9 10 7 5 4 3

Solution:-

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

 Income per day in â‚¹ Number of persons fi Mid â€“ points xi fixi |xi â€“ xÌ…| fi|xi â€“ xÌ…| 0 â€“ 100 4 50 200 308 1232 100 â€“ 200 8 150 1200 208 1664 200 â€“ 300 9 250 2250 108 972 300 â€“ 400 10 350 3500 8 80 400 â€“ 500 7 450 3150 92 644 500 â€“ 600 5 550 2750 192 960 600 â€“ 700 4 650 2600 292 1160 700 â€“ 800 3 750 2250 392 1176 50 17900 7896

10.

 Height in cms 95 â€“ 105 105 â€“ 115 115 â€“ 125 125 â€“ 135 135 â€“ 145 145 â€“ 155 Number of boys 9 13 26 30 12 10

Solution:-

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

 Height in cms Number of boys fi Mid â€“ points xi fixi |xi â€“ xÌ…| fi|xi â€“ xÌ…| 95 â€“ 105 9 100 900 25.3 227.7 105 â€“ 115 13 110 1430 15.3 198.9 115 â€“ 125 26 120 3120 5.3 137.8 125 â€“ 135 30 130 3900 4.7 141 135 â€“ 145 12 140 1680 14.7 176.4 145 â€“ 155 10 150 1500 24.7 247 100 12530 1128.8

11. Find the mean deviation about median for the following data:

 Marks 0 -10 10 -20 20 â€“ 30 30 â€“ 40 40 â€“ 50 50 â€“ 60 Number of girls 6 8 14 16 4 2

Solution:-

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

 Marks Number of Girls fi Cumulative frequency (c.f.) Mid â€“ points xi |xi â€“ Med| fi|xi â€“ Med| 0 â€“ 10 6 6 5 22.85 137.1 10 â€“ 20 8 14 15 12.85 102.8 20 â€“ 30 14 28 25 2.85 39.9 30 â€“ 40 16 44 35 7.15 114.4 40 â€“ 50 4 48 45 17.15 68.6 50 â€“ 60 2 50 55 27.15 54.3 50 517.1

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

So, 20-30 is the median class.

Then,

Median = l + (((N/2) â€“ c)/f) Ã— h

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

Median = 20 + (((25 â€“ 14))/14) Ã— 10

= 20 + 7.85

= 27.85

12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:

 Age (in years) 16 â€“ 20 21 â€“ 25 26 â€“ 30 31 â€“ 35 36 â€“ 40 41 â€“ 45 46 â€“ 50 51 â€“ 55 Number 5 6 12 14 26 12 16 9

[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Solution:-

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.

 Age Number fi Cumulative frequency (c.f.) Mid â€“ points xi |xi â€“ Med| fi|xi â€“ Med| 15.5 â€“ 20.5 5 5 18 20 100 20.5 â€“ 25.5 6 11 23 15 90 25.5 â€“ 30.5 12 23 28 10 120 30.5 â€“ 35.5 14 37 33 5 70 35.5 â€“ 40.5 26 63 38 0 0 40.5 â€“ 45.5 12 75 43 5 60 45.5 â€“ 50.5 16 91 48 10 160 50.5 â€“ 55.5 9 100 53 15 135 100 735

The class interval containing Nth/2 or 50th item is 35.5 â€“ 40.5

So, 35.5 â€“ 40.5 is the median class.

Then,

Median = l + (((N/2) â€“ c)/f) Ã— h

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

Median = 35.5 + (((50 â€“ 37))/26) Ã— 5

= 35.5 + 2.5

= 38

Exercise 15.2 Page: 371

Find the mean and variance for each of the data in Exercise 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12

Solution:-

So, xÌ… = (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.

 Xi Deviations from mean (xi â€“ xÌ…) (xi â€“ xÌ…)2 6 6 â€“ 9 = -3 9 7 7 â€“ 9 = -2 4 10 10 â€“ 9 = 1 1 12 12 â€“ 9 = 3 9 13 13 â€“ 9 = 4 16 4 4 â€“ 9 = â€“ 5 25 8 8 â€“ 9 = â€“ 1 1 12 12 â€“ 9 = 3 9 74

We know that Variance,

Ïƒ2 = (1/8) Ã— 74

= 9.2

âˆ´Mean = 9 and Variance = 9.25

2. First n natural numbers

Solution:-

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

âˆ´Mean, xÌ… = ((n(n + 1))2)/n

= (n + 1)/2

and also WKT Variance,

By substitute that value of xÌ… we get,

WKT, (a + b)(a â€“ b) = a2 â€“ b2

Ïƒ2 = (n2 â€“ 1)/12

âˆ´Mean = (n + 1)/2 and Variance = (n2 â€“ 1)/12

3. First 10 multiples of 3

Solution:-

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

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

So, xÌ… = (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.

 Xi Deviations from mean (xi â€“ xÌ…) (xi â€“ xÌ…)2 3 3 â€“ 16.5 = -13.5 182.25 6 6 â€“ 16.5 = -10.5 110.25 9 9 â€“ 16.5 = -7.5 56.25 12 12 â€“ 16.5 = -4.5 20.25 15 15 â€“ 16.5 = -1.5 2.25 18 18 â€“ 16.5 = 1.5 2.25 21 21 â€“ 16.5 = â€“ 4.5 20.25 24 24 â€“ 16.5 = 7.5 56.25 27 27 â€“ 16.5 = 10.5 110.25 30 30 â€“ 16.5 = 13.5 182.25 742.5

Then, Variance

= (1/10) Ã— 742.5

= 74.25

âˆ´Mean = 16.5 and Variance = 74.25

4.

 xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3

Solution:-

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

 Xi fi fixi Deviations from mean (xi â€“ xÌ…) (xi â€“ xÌ…)2 fi(xi â€“ xÌ…)2 6 2 12 6 â€“ 19 = 13 169 338 10 4 40 10 â€“ 19 = -9 81 324 14 7 98 14 â€“ 19 = -5 25 175 18 12 216 18 â€“ 19 = -1 1 12 24 8 192 24 â€“ 19 = 5 25 200 28 4 112 28 â€“ 19 = 9 81 324 30 3 90 30 â€“ 19 = 11 121 363 N = 40 760 1736

5.

 xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3

Solution:-

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

 Xi fi fixi Deviations from mean (xi â€“ xÌ…) (xi â€“ xÌ…)2 fi(xi â€“ xÌ…)2 92 3 276 92 â€“ 100 = -8 64 192 93 2 186 93 â€“ 100 = -7 49 98 97 3 291 97 â€“ 100 = -3 9 27 98 2 196 98 â€“ 100 = -2 4 8 102 6 612 102 â€“ 100 = 2 4 24 104 3 312 104 â€“ 100 = 4 16 48 109 3 327 109 â€“ 100 = 9 81 243 N = 22 2200 640

6. Find the mean and standard deviation using short-cut method.

 xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5

Solution:-

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

We obtain the following table from the given data.

 Xi Frequency fi Yi = (xi â€“ A)/h Yi2 fiyi fiyi2 60 2 -4 16 -8 32 61 1 -3 9 -3 9 62 12 -2 4 -24 48 63 29 -1 1 -29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 0 286

Mean,

Where A = 64, h = 1

So, xÌ… = 64 + ((0/100) Ã— 1)

= 64 + 0

= 64

Then, variance,

Ïƒ2 = (12/1002) [100(286) â€“ 02]

= (1/10000) [28600 â€“ 0]

= 28600/10000

= 2.86

Hence, standard deviation = Ïƒ = âˆš2.886

= 1.691

âˆ´ Mean = 64 and Standard Deviation = 1.691

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

7.

 Classes 0 â€“ 30 30 â€“ 60 60 â€“ 90 90 â€“ 120 120 â€“ 150 150 â€“ 180 180 â€“ 210 Frequencies 2 3 5 10 3 5 2

Solution:-

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

 Classes Frequency fi Mid â€“ points xi fixi (xi â€“ xÌ…) (xi â€“ xÌ…)2 fi(xi â€“ xÌ…)2 0 â€“ 30 2 15 30 -92 8464 16928 30 â€“ 60 3 45 135 -62 3844 11532 60 â€“ 90 5 75 375 -32 1024 5120 90 â€“ 120 10 105 1050 -2 4 40 120 â€“ 150 3 135 405 28 784 2352 150 â€“ 180 5 165 825 58 3364 16820 180 â€“ 210 2 195 390 88 7744 15488 N = 30 3210 68280

8.

 Classes 0 â€“ 10 10 â€“ 20 20 â€“ 30 30 â€“ 40 40 â€“50 Frequencies 5 8 15 16 6

Solution:-

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

 Classes Frequency fi Mid â€“ points xi fixi (xi â€“ xÌ…) (xi â€“ xÌ…)2 fi(xi â€“ xÌ…)2 0 â€“ 10 5 5 25 -22 484 2420 10 â€“ 20 8 15 120 -12 144 1152 20 â€“ 30 15 25 375 -2 4 60 30 â€“ 40 16 35 560 8 64 1024 40 â€“50 6 45 270 18 324 1944 N = 50 1350 6600

9. Find the mean, variance and standard deviation using short-cut method

 Height in cms 70 â€“ 75 75 â€“ 80 80 â€“ 85 85 â€“ 90 90 â€“ 95 95 â€“ 100 100 â€“ 105 105 â€“ 110 110 â€“ 115 Frequencies 3 4 7 7 15 9 6 6 3

Solution:-

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.

 Height (class) Number of children Frequency fi Midpoint Xi Yi = (xi â€“ A)/h Yi2 fiyi fiyi2 70 â€“ 75 3 72.5 -4 16 -12 48 75 â€“ 80 4 77.5 -3 9 -12 36 80 â€“ 85 7 82.5 -2 4 -14 28 85 â€“ 90 7 87.5 -1 1 -7 7 90 â€“ 95 15 92.5 0 0 0 0 95 â€“ 100 9 97.5 1 1 9 9 100 â€“ 105 6 102.5 2 4 12 24 105 â€“ 110 6 107.5 3 9 18 54 110 â€“ 115 3 112.5 4 16 12 48 N = 60 6 254

Mean,

Where, A = 92.5, h = 5

So, xÌ… = 92.5 + ((6/60) Ã— 5)

= 92.5 + Â½

= 92.5 + 0.5

= 93

Then, Variance,

Ïƒ2 = (52/602) [60(254) â€“ 62]

= (1/144) [15240 â€“ 36]

= 15204/144

= 1267/12

= 105.583

Hence, standard deviation = Ïƒ = âˆš105.583

= 10.275

âˆ´ Mean = 93, variance = 105.583 and Standard Deviation = 10.275

10. The diameters of circles (in mm) drawn in a design are given below:

 Diameters 33 â€“ 36 37 â€“ 40 41 â€“ 44 45 â€“ 48 49 â€“ 52 No. of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.

[Hint first make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 â€“ 48.5, 48.5 â€“ 52.5 and then proceed.]

Solution:-

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.

 Height (class) Number of children (Frequency fi) Midpoint Xi Yi = (xi â€“ A)/h Yi2 fiyi fiyi2 32.5 â€“ 36.5 15 34.5 -2 4 -30 60 36.5 â€“ 40.5 17 38.5 -1 1 -17 17 40.5 â€“ 44.5 21 42.5 0 0 0 0 44.5 â€“ 48.5 22 46.5 1 1 22 22 48.5 â€“ 52.5 25 50.5 2 4 50 100 N = 100 25 199

Mean,

Where, A = 42.5, h = 4

So, xÌ… = 42.5 + (25/100) Ã— 4

= 42.5 + 1

= 43.5

Then, Variance,

Ïƒ2 = (42/1002)[100(199) â€“ 252]

= (1/625) [19900 â€“ 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = Ïƒ = âˆš30.84

= 5.553

âˆ´ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.

Exercise 15.3 Page: 375

1. From the data given below state which group is more variable, A or B?

 Marks 10 â€“ 20 20 â€“ 30 30 â€“ 40 40 â€“ 50 50 â€“ 60 60 â€“ 70 70 â€“ 80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7

Solution:-

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.) = (Ïƒ/ xÌ…) Ã— 100

Where, Ïƒ = standard deviation, xÌ… = mean

For Group A

 Marks Group A fi Mid-point Xi Yi = (xi â€“ A)/h (Yi)2 fiyi fi(yi)2 10 â€“ 20 9 15 ((15 â€“ 45)/10) = -3 (-3)2= 9 â€“ 27 81 20 â€“ 30 17 25 ((25 â€“ 45)/10) = -2 (-2)2= 4 â€“ 34 68 30 â€“ 40 32 35 ((35 â€“ 45)/10) = â€“ 1 (-1)2= 1 â€“ 32 32 40 â€“ 50 33 45 ((45 â€“ 45)/10) = 0 02 0 0 50 â€“ 60 40 55 ((55 â€“ 45)/10) = 1 12= 1 40 40 60 â€“ 70 10 65 ((65 â€“ 45)/10) = 2 22= 4 20 40 70 â€“ 80 9 75 ((75 â€“ 45)/10) = 3 32= 9 27 81 Total 150 -6 342

Where A = 45,

and yi = (xi â€“ A)/h

Here h = class size = 20 â€“ 10

h = 10

So, xÌ… = 45 + ((-6/150) Ã— 10)

= 45 â€“ 0.4

= 44.6

Ïƒ2 = (102/1502) [150(342) â€“ (-6)2]

= (100/22500) [51,300 â€“ 36]

= (100/22500) Ã— 51264

= 227.84

Hence, standard deviation = Ïƒ = âˆš227.84

= 15.09

âˆ´ C.V for group A = (Ïƒ/ xÌ…) Ã— 100

= (15.09/44.6) Ã— 100

= 33.83

Now, for group B.

 Marks Group B fi Mid-point Xi Yi = (xi â€“ A)/h (Yi)2 fiyi fi(yi)2 10 â€“ 20 10 15 ((15 â€“ 45)/10) = -3 (-3)2= 9 â€“ 30 90 20 â€“ 30 20 25 ((25 â€“ 45)/10) = -2 (-2)2= 4 â€“ 40 80 30 â€“ 40 30 35 ((35 â€“ 45)/10) = â€“ 1 (-1)2= 1 â€“ 30 30 40 â€“ 50 25 45 ((45 â€“ 45)/10) = 0 02 0 0 50 â€“ 60 43 55 ((55 â€“ 45)/10) = 1 12= 1 43 43 60 â€“ 70 15 65 ((65 â€“ 45)/10) = 2 22= 4 30 60 70 â€“ 80 7 75 ((75 â€“ 45)/10) = 3 32= 9 21 63 Total 150 -6 366

Where A = 45,

h = 10

So, xÌ… = 45 + ((-6/150) Ã— 10)

= 45 â€“ 0.4

= 44.6

Ïƒ2 = (102/1502) [150(366) â€“ (-6)2]

= (100/22500) [54,900 â€“ 36]

= (100/22500) Ã— 54,864

= 243.84

Hence, standard deviation = Ïƒ = âˆš243.84

= 15.61

âˆ´ C.V for group B = (Ïƒ/ xÌ…) Ã— 100

= (15.61/44.6) Ã— 100

= 35

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

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

So, Group B is more variable.

2. From the prices of shares X and Y below, find out which is more stable in value:

 X 35 54 52 53 56 58 52 50 51 49 Y 108 107 105 105 106 107 104 103 104 101

Solution:-

From the given data,

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

 X (xi) Y (yi) Xi2 Yi2 35 108 1225 11664 54 107 2916 11449 52 105 2704 11025 53 105 2809 11025 56 106 8136 11236 58 107 3364 11449 52 104 2704 10816 50 103 2500 10609 51 104 2601 10816 49 101 2401 10201 Total = 510 1050 26360 110290

We have to calculate Mean for x,

Mean xÌ… = âˆ‘xi/n

Where, n = number of terms

= 510/10

= 51

Then, Variance for x =

= (1/102)[(10 Ã— 26360) â€“ 5102]

= (1/100) (263600 â€“ 260100)

= 3500/100

= 35

WKT Standard deviation = âˆšvariance

= âˆš35

= 5.91

So, co-efficient of variation = (Ïƒ/ xÌ…) Ã— 100

= (5.91/51) Ã— 100

= 11.58

Now, we have to calculate Mean for y,

Mean È³ = âˆ‘yi/n

Where, n = number of terms

= 1050/10

= 105

Then, Variance for y =

= (1/102)[(10 Ã— 110290) â€“ 10502]

= (1/100) (1102900 â€“ 1102500)

= 400/100

= 4

WKT Standard deviation = âˆšvariance

= âˆš4

= 2

So, co-efficient of variation = (Ïƒ/ xÌ…) Ã— 100

= (2/105) Ã— 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.

3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

 Firm A Firm B No. of wages earners 586 648 Mean of monthly wages Rs 5253 Rs 5253 Variance of the distribution of wages 100 121

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

Solution:-

(i) From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 Ã— 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 Ã— 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 (Ïƒ)= âˆš100

=10

Variance of firm B = 121

Then,

Standard deviation (Ïƒ)=âˆš(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.

4. The following is the record of goals scored by team A in a football session:

 No. of goals scored 0 1 2 3 4 No. of matches 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Solution:-

From the given data,

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

 Number of goals scored xi Number of matches fi fixi Xi2 fixi2 0 1 0 0 0 1 9 9 1 9 2 7 14 4 28 3 5 15 9 45 4 3 12 16 48 Total 25 50 130

Since C.V. of firm B is greater

âˆ´ Team A is more consistent.

5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

Which is more varying, the length or weight?

Solution:-

First we have to calculate Mean for Length x,

Â

Miscellaneous Exercise Page: 380

1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Solution:-

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Solution:-

3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Solution:-

We know that,

4. Given that xÌ… is the mean and Ïƒ2 is the variance of n observations x1, x2, â€¦,xn . Prove that the mean and variance of the observations ax1, ax2, ax3, â€¦., axn are axÌ… and a2Ïƒ2, respectively, (a â‰  0).

Solution:-

From the question it is given that, n observations are x1, x2,â€¦..xn

Mean of the n observation = xÌ…

Variance of the n observation = Ïƒ2

As we know that,

5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12

Solution:-

(i) If 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

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(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 â€“ 8 + 12

n = 204

Then, correct mean = correct sum/20

= 204/20

= 10.2

6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

 Subject Mathematics Physics Chemistry Mean 42 32 40.9 Standard deviation 12 15 20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Solution:-

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 that,

7. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Solution:-

From the question it is given that,

The total number of observations (n) = 100

Incorrect mean, (xÌ…) = 20

And, Incorrect standard deviation (Ïƒ) = 3

 Also Access NCERT Exemplar for Class 11 Maths Chapter 15 CBSE Notes for Class 11 Maths Chapter 15

NCERT Solutions for Class 11 Maths Chapter 15 â€“ Statistics

The topics of Class 11 Chapter 15 â€“ Statistics of NCERT Solution are as follows:

15.1 Introduction

This section talks about the concept of central tendency, mean and median [during even and the odd number of observations] with examples. It introduces the concept of measure of dispersion.

The values which cluster around the middle or centre of the distribution are measures of central tendency. They are mean, median and mode.

In a class, mean can be used to find the average marks scored by the students.

When calculating the height of students, the median can be used to find the middlemost value.

15.2 Measures of Dispersion

This section defines measures of dispersion, different measures of dispersion [range, quartile deviation, mean deviation, standard deviation]

Measures of dispersion explain the relationship with measures of central tendency. For example, the spread of data tells how well the mean represents the data. If the spread is large, then mean is not representative of data.

15.3 Range

This section defines the range, its formula and an example.

The range gives the variability of scores using the maximum and minimum values in the set.

In a cricket match, Batsman A range = 121 â€“ 0 = 121 and Batsman B range = 52 â€“ 22 = 30

Range A > Range B. So, the scores are spread in case of A whereas for B they are close to each other.

15.4 Mean Deviation

This section defines mean deviation, the formula for mean deviation.

The concept of mean deviation can be used by biologists in the comparison of different animal weights and decide what would be a healthy weight.

15.4.1Mean deviation for ungrouped data

The process of obtaining the mean deviation for ungrouped data is elaborated in this section.

Find the mean, deviations from the mean, absolute deviations and substitute the values in the mean deviation formula and arrive at the answer.

15.4.2 Mean deviation for grouped data

The process of obtaining mean deviation for discrete and continuous frequency distributions are elaborated in this section with solved examples.

15.4.3 Limitations of mean deviation

• If in a series the degree of variability is very high, then median will not be a representative of data. Hence the mean deviation calculated about such median cannot be relied on.
• If the sum of deviations from the mean is greater than the sum of deviations from the median, then the mean deviation about mean is not very specific.
• The absolute mean deviation calculated canâ€™t be subjected to further algebraic treatment. It canâ€™t be used as an appropriate measure of dispersion.

15.5 Variance and standard deviation

15.5.1 Standard Deviation

This section involves the variance and standard deviation definition, formula and solved examples on the discrete and continuous frequency distribution.

A science test was taken by a class of students. The mean score of the test was 85% on calculation. The teacher found the standard deviation of other scores and noticed that a very small standard deviation exists which suggests that most of the students scored very close to 85%.

15.5.2 Shortcut method to find Variance and standard deviation

This section deals with the simpler way of calculating the standard deviation with a few illustrations.

15.6 Analysis of Frequency Distributions

This section deals with the process of comparing the variability of two series having the same mean, coefficient of variation with few solved problems.

A few points on Chapter 15 Statistics

• Range is defined as the difference between the maximum and minimum value of the given data.
• If there exists series with equal means, then the series with lesser standard deviation is more consistent or less scattered.
• The addition or subtraction of a positive number to each data point of the data set will not affect the variance.

The solutions give substitute strategies and clarifications to take care of problems which makes the student feel sure while taking the exam. Additionally, taking care of many muddled issues upgrades the information on Mathematical abilities in the students. The solutions cater to all the vital inquiries which a student should and ought to have aced to show up for the exam. The BYJUâ€™S subject specialists who have written the NCERT Solutions for Class 11 Maths according to the latest CBSE Syllabus 2022-23 have complete knowledge about the question paper setting and the marks distributed across the chapters.

Disclaimer â€“Â

Dropped Topics â€“Â

15.6 Analysis of Frequency Distribution
Ques. 6 (Miscellaneous Exercise) and last point in the Summary

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 15

Explain the term standard deviation which is discussed in the Chapter 15 of the NCERT Solutions for Class 11 Maths.

Standard deviation deals with the measurement of variation or deviation of the given set of values. The range is determined based on the level of standard deviation. To understand this term in a better way, students are advised to download the PDF of solutions available at BYJUâ€™S. The solutions are present in chapter wise and exercise wise format based on the requirement of students.

What are the topics discussed in the Chapter 15 of NCERT Solutions for Class 11 Maths?

The topics discussed in the Chapter 15 of NCERT Solutions for Class 11 Maths are â€“
1. Introduction
2. Methods of Dispersion
3. Range
4. Mean Deviation
5. Variance and Standard Deviation
6. Analysis of Frequency Distributions

Are the NCERT Solutions for Class 11 Maths Chapter 15 helpful for the students?

The NCERT Solutions for Class 11 Maths Chapter 15 provides accurate explanations in simple language to help students score well in the exams. The step by step method of solving problems provides a clear idea to the students about the marks weightage as per the updated CBSE Syllabus 2022-23. Students will be able to analyse their areas of weakness and work on them for a better academic score.