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**NCERT Class 11 Mathematics Chapter 13 Statistics**

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**Solutions****NCERT Class 11 Mathematics Chapter 13 Statistics****Statistics**

**Statistics****Chapter – 13**

Exercise 13.1 |

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

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

Ans: The given data is

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

Mean of the data,

The deviations of the respective observations from the mean X̅, i.e. xᵢ – X̅ , are

– 6, – 3, – 2, – 1, 0, 2, 3, 7

The absolute values of the deviations, i.e. | xᵢ – X̅ | are

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

The required mean deviation about the mean is

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

Ans: The given data is

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

Mean of the given data,

The deviations of the respective observations from the mean X̅ , i.e. xᵢ – X̅ , are

– 12,20, – 2,- 10,- 8,5,13,- 4,4,- 6

The absolute values of the deviations, i.e.| xᵢ – X̅ | are

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

The required mean deviation about the mean is

**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.**

Ans: The given data is

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

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

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

The deviations of the respective observations 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

The absolute values of the deviations, |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

The required mean deviation about the median is

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

Ans: The given data is

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

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

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

The deviations of the respective observations 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

The absolute values of the deviations, |xᵢ – M|, are

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

Thus, the required mean deviation about the median is

**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 |

Ans:

xᵢ | fᵢ | fᵢ xᵢ | |xᵢ – X̅| | fᵢ|xᵢ – X̅| |

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 |

**6.**

xᵢ | 10 | 30 | 50 | 70 | 90 |

fᵢ | 4 | 24 | 28 | 16 | 8 |

Ans:

xᵢ | fᵢ | fᵢ xᵢ | |xᵢ – X̅| | fᵢ|xᵢ – X̅| |

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 |

Ans: The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xᵢ | fᵢ | c.f. |

5 | 8 | 8 |

7 | 6 | 14 |

9 | 2 | 16 |

10 | 2 | 18 |

12 | 2 | 20 |

15 | 6 | 26 |

Here, N = 26, which is even.

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

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

|xᵢ â€” M| | 2 | 0 | 2 | 3 | 5 | 8 |

fᵢ | 8 | 6 | 2 | 2 | 2 | 6 |

fᵢ|xᵢ â€” M| | 16 | 0 | 4 | 6 | 10 | 48 |

**8.**

xᵢ | 15 | 21 | 27 | 30 | 35 |

fᵢ | 3 | 5 | 6 | 7 | 8 |

Ans: The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

xᵢ | fᵢ | c.f. |
---|---|---|

15 | 3 | 3 |

21 | 5 | 8 |

27 | 6 | 14 |

30 | 7 | 21 |

35 | 8 | 29 |

Here, N = 29, which is odd.

observation = 15ᵗʰ observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

.: Median = 30

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

|xᵢ – M| | 15 | 9 | 3 | 0 | 5 |

fᵢ | 3 | 5 | 6 | 7 | 8 |

fᵢ|xᵢ – M| | 45 | 45 | 18 | 0 | 40 |

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

**9.**

Income per day in ₹ | 0 -100 | 100 – 200 | 200 – 300 | 300 – 400 | 400 – 500 | 500 – 600 | 600 – 700 | 700 – 800 |

Number of persons | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |

Ans: The following table is formed.

Income per day in | Number of persons fᵢ | Mid – point xᵢ | fᵢ xᵢ | |xᵢ – X̅| | fᵢ|xᵢ – X̅| |

0 -100 | 4 | 50 | 200 | 308 | 1232 |

100 – 200 | 8 | 150 | 1200 | 208 | 1664 |

200 – 300 | 9 | 250 | 2250 | 108 | 972 |

300 – 400 | 10 | 350 | 3500 | 8 | 80 |

400 – 500 | 7 | 450 | 3150 | 92 | 644 |

500 – 600 | 5 | 550 | 2750 | 192 | 960 |

600 – 700 | 4 | 650 | 2600 | 292 | 1168 |

700 – 800 | 3 | 750 | 2250 | 392 | 1176 |

50 | 17900 | 7896 |

**10. **

Height in cms | 95 – 105 | 105 – 115 | 115 – 125 | 125 – 135 | 135 – 145 | 145 – 155 |

Number of boys | 9 | 13 | 26 | 30 | 12 | 10 |

Ans: The following table is formed.

Height in cms | Number of boys fᵢ | Mid – point xᵢ | fᵢ xᵢ | |xᵢ – X̅| | fᵢ|xᵢ – X̅| |

95 – 105 | 9 | 100 | 900 | 25.3 | 227.7 |

105 – 115 | 13 | 110 | 1430 | 15.3 | 198.9 |

115 – 125 | 26 | 120 | 3120 | 5.3 | 137.8 |

125 – 135 | 30 | 130 | 3900 | 4.7 | 141 |

135 – 145 | 12 | 140 | 1680 | 14.7 | 176.4 |

145 – 155 | 10 | 150 | 1500 | 24.7 | 247 |

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

Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30-40 | 40-50 | 50-60 |

Number of girls | 6 | 8 | 14 | 16 | 4 | 2 |

Ans:

Marks | fᵢ | Cumulative frequency | xᵢ | |xᵢ – M| | fᵢ|xᵢ – M| |

0 – 10 | 6 | 6 | 5 | 22.86 | 137.16 |

10 – 20 | 8 | 14 | 15 | 12.86 | 102.88 |

20 – 30 | 14 | 28 | 25 | 2.86 | 40.04 |

30 – 40 | 16 | 44 | 35 | 7.14 | 114.24 |

40 – 50 | 4 | 48 | 45 | 17.14 | 68.56 |

50 – 60 | 2 | 50 | 55 | 27.14 | 54.28 |

Total | 50 | 517.32 |

Since, 25 occurs in the cumulative frequency 25.

So, the median class is 20 – 30.

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

Age ( in years) | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 36 – 40 | 41 – 45 | 46 – 50 | 51 – 55 |

Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |

Ans: The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by

subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age | Number fᵢ | Cumulative frequency(c.f.) | Mid – point xᵢ | |xᵢ – Med.| | fᵢ|xᵢ – Med.| |

15.5 – 20.5 | 5 | 5 | 18 | 20 | 100 |

20.5 – 25.5 | 6 | 11 | 23 | 15 | 90 |

25.5 – 30.5 | 12 | 23 | 28 | 10 | 120 |

30.5 – 35.5 | 14 | 37 | 33 | 5 | 70 |

35.5 – 40.5 | 26 | 63 | 38 | 0 | 0 |

40.5 – 45.5 | 12 | 75 | 43 | 5 | 60 |

45.5 – 50.5 | 16 | 91 | 48 | 10 | 160 |

50.5 – 55.5 | 9 | 100 | 53 | 15 | 135 |

100 | 735 |

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

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, /= 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, mean deviation about the median is given by,

Exercise 13.2 |

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

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

Ans: 6, 7, 10, 12, 13, 4, 8, 12

Mean,

The following table is obtained.

xᵢ | (xᵢ – X̅) | (xᵢ – X̅)² |

6 | – 3 | 9 |

7 | – 2 | 4 |

10 | -1 | 1 |

12 | 3 | 9 |

13 | 4 | 16 |

4 | – 5 | 25 |

8 | – 1 | 1 |

12 | 3 | 9 |

74 |

**2. First n natural numbers.**

Ans: The mean of first n natural numbers is calculated as follows.

**3. First 10 multiples of 3.**

Ans: The first 10 multiples of 3 are

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

Here, number of observations, n = 10

The following table is obtained.

xᵢ | (xᵢ – X̅) | (xᵢ – X̅)² |

3 | – 13.5 | 182.25 |

6 | -10.5 | 110.25 |

9 | – 7.5 | 56.25 |

12 | – 4.5 | 20.25 |

15 | – 1.5 | 2.25 |

18 | 1.5 | 2.25 |

21 | 4.5 | 20.25 |

24 | 7.5 | 56.25 |

27 | 10.5 | 110.25 |

30 | 13.5 | 182.25 |

742.5 |

**4.**

xᵢ | 6 | 10 | 14 | 18 | 24 | 28 | 30 |

fᵢ | 2 | 4 | 7 | 12 | 8 | 4 | 3 |

Ans: The data is obtained in tabular form as follows.

xᵢ | fᵢ | fᵢxᵢ | xᵢ – X̅ | (xᵢ – X̅)² | fᵢ(xᵢ – X̅)² |

6 | 2 | 12 | -13 | 169 | 338 |

10 | 4 | 40 | -9 | 81 | 324 |

14 | 7 | 98 | -5 | 25 | 175 |

18 | 12 | 216 | -1 | 1 | 12 |

24 | 8 | 192 | 5 | 25 | 200 |

28 | 4 | 112 | 9 | 81 | 324 |

30 | 3 | 90 | 11 | 121 | 363 |

40 | 760 | 1736 |

**5.**

xᵢ | 92 | 93 | 97 | 98 | 102 | 104 | 109 |

fᵢ | 3 | 2 | 3 | 2 | 6 | 3 | 3 |

Ans: The data is obtained in tabular form as follows.

xᵢ | fᵢ | fᵢxᵢ | xᵢ – X̅ | (xᵢ – X̅)² | fᵢ(xᵢ – X̅)² |

92 | 3 | 276 | -8 | 64 | 192 |

93 | 2 | 186 | -7 | 49 | 98 |

97 | 3 | 291 | -3 | 9 | 27 |

98 | 2 | 196 | -2 | 4 | 8 |

102 | 6 | 612 | 2 | 4 | 24 |

104 | 3 | 312 | 4 | 16 | 48 |

109 | 3 | 327 | 9 | 81 | 243 |

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 |

Ans: The data is obtained in tabular form as follows.

xᵢ | fᵢ | 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 |

100 | 220 | 0 | 286 |

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

**7.**

Classes | 0 – 30 | 30 – 60 | 60 – 90 | 90 – 120 | 120 – 150 | 150 – 180 | 180 – 210 |

Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |

Ans:

Class | Frequencyfᵢ | Mid – point xᵢ | yᵢ² | fyᵢ | fyᵢ² | |

0 – 30 | 2 | 15 | -3 | 9 | -6 | 18 |

30 – 60 | 3 | 45 | -2 | 4 | -6 | 12 |

60 – 90 | 5 | 75 | -1 | 1 | -5 | 5 |

90 – 120 | 10 | 105 | 0 | 0 | 0 | 0 |

120 – 150 | 3 | 135 | 1 | 1 | 3 | 3 |

150 – 180 | 5 | 165 | 2 | 4 | 10 | 20 |

180 – 210 | 2 | 195 | 3 | 9 | 6 | 18 |

30 | 2 | 76 |

**8.**

Classes | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |

Frequencies | 5 | 8 | 15 | 16 | 6 |

Ans:

Class | Frequencyfᵢ | Mid – point xᵢ | yᵢ² | fyᵢ | fyᵢ² | |

0 – 10 | 5 | 5 | -2 | 4 | -10 | 20 |

10 – 20 | 8 | 15 | -1 | 1 | -8 | 8 |

20 – 30 | 15 | 25 | 0 | 0 | 0 | 0 |

30 – 40 | 16 | 35 | 1 | 1 | 16 | 16 |

40 – 50 | 6 | 45 | 2 | 4 | 12 | 24 |

50 | 10 | 68 |

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

Heightin cms | 70-75 | 75-80 | 80-85 | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |

No. ofchildren | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |

Ans:

Class Interval | Frequencyfᵢ | Mid – point xᵢ | yᵢ² | fyᵢ | fyᵢ² | |

70 – 75 | 3 | 72.5 | -4 | 16 | -12 | 48 |

75 – 80 | 4 | 77.5 | -3 | 9 | -12 | 36 |

80 – 85 | 7 | 82.5 | -2 | 4 | -14 | 28 |

85 – 90 | 7 | 87.5 | -1 | 1 | -7 | 7 |

90 – 95 | 15 | 92.5 | 0 | 0 | 0 | 0 |

95 -100 | 9 | 97.5 | 1 | 1 | 9 | 9 |

100 -105 | 6 | 102.5 | 2 | 4 | 12 | 24 |

105 -110 | 6 | 107.5 | 3 | 9 | 18 | 54 |

110 – 115 | 3 | 112.5 | 4 | 16 | 12 | 48 |

60 | 6 | 254 |

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

Diameters | 33 – 36 | 37 – 40 | 41 – 44 | 45 – 48 | 49 – 52 |

No. of circles | 15 | 17 | 21 | 22 | 25 |

Ans:

Class Interval | Frequencyfᵢ | Mid – point xᵢ | fᵢ² | fyᵢ | fyᵢ² | |

32.5 – 36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |

36.5 – 40.5 | 17 | 38.5 | -1 | 1 | -17 | 17 |

40.5 – 44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |

44.5 – 48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |

48.5 – 52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |

100 | 25 | 199 |

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.

Miscellaneous Exercise on Chapter 12 |

**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. **

Ans: Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, х, у.

⇒ 60 + x + y = 72

⇒ x + y = 12 …(1)

⇒ x² + y² = 80 …(2)

From (1), we obtain

x² + y² + 2xy =144 -| (3)

From (2) and (3), we obtain

2xy = 64 -| (4)

Subtracting (4) from (2), we obtain

x² + y² – 2xy = 80 – 64 = 16

⇒ x – y= A ± 4-|(5)

Therefore, from (1) and (5), we obtain

x = 8 and Y = 4 when x – Y = 4

x = 4 and y = 8 when x – y = – 4

Thus, the remaining observations are 4 and 8.

**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. **

Ans: Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

⇒ 56 = 42 + x + y

⇒x + y = 14 ….(1)

⇒ x² + y² = 112 – 12 = 100

x² + y² = 100 …..(2)

From (1), we obtain

x² + y² + 2xy = 196 -| (3)

From (2) and (3), we obtain

2xy =196 – 100

⇒ 2xy = 96 -| (4)

Subtracting (4) from (2), we obtain

x² + y² – 2xy = 100 – 96

⇒ (x – y)² = 4

⇒ x – y= A ± 2 -| (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 6 when x – y = 2

x = 6 and y = 8 when x – y = – 2

Thus, the remaining observations are 6 and 8.

**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.**

Ans: Let the observations be X₁ , X₂ , X₃ , X₄ ,X₅ , and X₆ .

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are y, then

= 3 × 8 ….[Using(1)]

= 24

Standard deviation,

From (1) and (2), it can be observed that,

Substituting the values of xᵢ and X̅ in (2), we obtain.

Therefore, variance of new observations =

Hence, the standard deviation of new observations is √144 = 12

**4. Given that X̅ is the mean and Ïf² is the variance of n observations x₁, x₂ -| xₙ.Prove that the mean and variance of the observations ax₁, ax₂, ax₃ -axₙ are aX̅ and a² Ïf², respectively, ( a ≠ 0).**

Ans: The given n observations are x₁ ,x₂ -| xₙ.

Mean = X̅

Variance = Ïf²

If each observation is multiplied by a and the new observations are y, then

Therefore, mean of the observations, ax₁, ax₂ -| axₙ, is aX̅.

Substituting the values of xand X̅ in (1), we obtain

Thus, the variance of the observations, ax₁,ax₂ -| axₙ, is a² Ïf² .

**Q5: 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 the wrong item is omitted.**

Ans: Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

∴Correct mean

**(ii) If it is replaced by 12.**

Ans: When 8 is replaced by 12,

Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

**6. 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.**

Ans: Number of observations (n) = 100

Incorrect mean ( X̅ )=20

Incorrect standard deviation (σ) = 3

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940