# NCERT Solutions for Class 10th: Ch 14 Statistics Maths

# NCERT Solutions for Class 10th: Ch 14 Statistics Maths

#### NCERT Solutions for Class 10th: Ch 14 Statistics Maths

Page No: 270**Exercise 14.1**

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |

Number of Houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |

**Answer**

No. of plants (Class interval) | No. of houses (f_{i}) | Mid-point (x_{i}) | f_{i}x_{i } |

0-2 | 1 | 1 | 1 |

2-4 | 2 | 3 | 6 |

4-6 | 1 | 5 | 5 |

6-8 | 5 | 7 | 35 |

8-10 | 6 | 9 | 54 |

10-12 | 2 | 11 | 22 |

12-14 | 3 | 13 | 39 |

Sum f_{i }= 20_{} | Sum f_{i}x_{i} = 162 |

Mean = x̄ = ∑f

_{i}x

_{i}/∑f

_{i }= 162/20 = 8.1

_{}

We would use direct method because the numerical value of f

_{i}and x

_{i}are small.

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |

Number of workers | 12 | 14 | 8 | 6 | 10 |

**Answer**

Here, the value of mid-point (x

_{i}) is very large, so assumed mean A = 150 and class interval is h = 20.

So, u

_{i }= (x

_{i}- A)/h = u

_{i }= (x

_{i}- 150)/20

Daily wages (Class interval) | Number of workers frequency (f _{i}) | Mid-point (x_{i}) | u_{i }= (x_{i} - 150)/20 | f_{i}u_{i } |

100-120 | 12 | 110 | -2 | -24 |

120-140 | 14 | 130 | -1 | -14 |

140-160 | 8 | 150 | 0 | 0 |

160-180 | 6 | 170 | 1 | 6 |

180-200 | 10 | 190 | 2 | 20 |

Total | Sum f_{i }= 50_{} | Sum f_{i}u_{i} = -12 |

_{i}u

_{i}/∑f

_{i }=150 + (20 × -12/50) = 150 - 4.8 = 145.20

Thus, mean daily wage = Rs. 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

**Answer**

Here, the value of mid-point (x

_{i}) mean x̄ = 18

Class interval | Number of children (f_{i}) | Mid-point (x_{i}) | f_{i}x_{i } |

11-13 | 7 | 12 | 84 |

13-15 | 6 | 14 | 84 |

15-17 | 9 | 16 | 144 |

17-19 | 13 | 18 = A | 234 |

19-21 | f | 20 | 20f |

21-23 | 5 | 22 | 110 |

23-25 | 4 | 24 | 96 |

Total | f_{i} = 44+f_{} | Sum f_{i}x_{i} = 752+20f |

Mean = x̄ = ∑f

_{i}x

_{i}/∑f

_{i }= (752+20f)/(44+f)

⇒ 18 = (752+20f)/(44+f)

⇒ 18(44+f) = (752+20f)

⇒ 792+18f = 752+20f

⇒ 792+18f = 752+20f

⇒ 792 - 752 = 20f - 18f

⇒ 40 = 2f

⇒ f = 20

Page No: 271

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

**Answer**

x

_{i }= (Upper limit + Lower limit)/2

Class size (h) = 3

Assumed mean (A) = 75.5

Class Interval | Number of women (f_{i}) | Mid-point (x_{i}) | u_{i} = (x_{i} - 75.5)/h | f_{i}u_{i} |

65-68 | 2 | 66.5 | -3 | -6 |

68-71 | 4 | 69.5 | -2 | -8 |

71-74 | 3 | 72.5 | -1 | -3 |

74-77 | 8 | 75.5 | 0 | 0 |

77-80 | 7 | 78.5 | 1 | 7 |

80-83 | 4 | 81.5 | 3 | 8 |

83-86 | 2 | 84.5 | 3 | 6 |

Sum f_{i}= 30 | Sum f_{i}u_{i }= 4 |

Mean = x̄ = A + h∑f

_{i}u

_{i}/∑f

_{i }= 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9

The mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

**Answer**

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.

Here, assumed mean (A) = 57

Class size (h) = 3

Class Interval | Number of boxes (f_{i}) | Mid-point (x_{i}) | d_{i} = x_{i} - A | f_{i}d_{i} |

49.5-52.5 | 15 | 51 | -6 | 90 |

52.5-55.5 | 110 | 54 | -3 | -330 |

55.5-58.5 | 135 | 57 = A | 0 | 0 |

58.5-61.5 | 115 | 60 | 3 | 345 |

61.5-64.5 | 25 | 63 | 6 | 150 |

Sum f_{i} = 400 | Sum f_{i}d_{i} = 75 |

Mean = x̄ = A + ∑f

_{i}d

_{i}/∑f

_{i }= 57 + (75/400) = 57 + 0.1875 = 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

**Answer**

Here, assumed mean (A) = 225

Class Interval | Number of households (f_{i}) | Mid-point (x_{i}) | d_{i} = x_{i} - A | f_{i}d_{i} |

100-150 | 4 | 125 | -100 | -400 |

150-200 | 5 | 175 | -50 | -250 |

200-250 | 12 | 225 | 0 | 0 |

250-300 | 2 | 275 | 50 | 100 |

300-350 | 2 | 325 | 100 | 200 |

Sum f_{i} = 25 | Sum f_{i}d_{i} = -350 |

Mean = x̄ = A + ∑f

_{i}d

_{i}/∑f

_{i }= 225 + (-350/25) = 225 - 14 = 211

The mean daily expenditure on food is 211

7. To find out the concentration of SO

_{2}in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

_{2}in the air.

**Answer**

Concentration of SO_{2 }(in ppm) | Frequency (f_{i}) | Mid-point (x_{i}) |
f
_{i}x_{i} |

0.00-0.04 | 4 | 0.02 | 0.08 |

0.04-0.08 | 9 | 0.06 | 0.54 |

0.08-0.12 | 9 | 0.10 | 0.90 |

0.12-0.16 | 2 | 0.14 | 0.28 |

0.16-0.20 | 4 | 0.18 | 0.72 |

0.20-0.24 | 2 | 0.20 | 0.40 |

Total | Sum f_{i} = 30 | Sum (f_{i}x_{i}) = 2.96 |

Mean = x̄ = ∑f

_{i}x

_{i}/∑f

_{i}

= 2.96/30 = 0.099 ppm

Page No. 272

8. A class teacher has the following absentee record of 40 students of a class for the whole

term. Find the mean number of days a student was absent.

Number of days
| 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
---|---|---|---|---|---|---|---|

Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |

**Answer**

Class interval | Frequency (f_{i}) |
Mid-point (x
_{i}) |
f
_{i}x_{i} |

0-6 | 11 | 3 | 33 |

6-10 | 10 | 8 | 80 |

10-14 | 7 | 12 | 84 |

14-20 | 4 | 17 | 68 |

20-28 | 4 | 24 | 96 |

28-38 | 3 | 33 | 99 |

38-40 | 1 | 39 | 39 |

Sum f_{i} = 40 | Sum f_{i}x_{i} = 499 |

Mean = x̄ = ∑f

_{i}x

_{i}/∑f

_{i}

= 499/40 = 12.48 days

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean

literacy rate.

Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-98 |

Number of cities | 3 | 10 | 11 | 8 | 3 |

**Answer**

Class Interval | Frequency (f_{i}) | (x_{i}) | d_{i} = x_{i} - a | u_{i} = d_{i}/h |
f
_{i}u_{i} |

45-55 | 3 | 50 | -20 | -2 | -6 |

55-65 | 10 | 60 | -10 | -1 | -10 |

65-75 | 11 | 70 | 0 | 0 | 0 |

75-85 | 8 | 80 | 10 | 1 | 8 |

85-95 | 3 | 90 | 20 | 2 | 6 |

Sum f_{i} = 35 | Sum f_{i}u_{i} = -2 |

Mean = x̄ = a + (∑f

_{i}u

_{i}/∑f

_{i}) х h

= 70 + (-2/35) х 10 = 69.42

Page No. 275

**Exercise 14.2**

1. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |

Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |

Find the mode and the mean of the data given above. Compare and interpret the two

measures of central tendency.

**Answer**

Modal class = 35 – 45, l = 35, class width (h) = 10, f

_{m}= 23, f

_{1}= 21 and f

_{2}= 14

Calculation of Mean:

Class Interval | Frequency (f_{i}) | Mid-point (x_{i}) | f_{i}x_{i} |

5-15 | 6 | 10 | 60 |

15-25 | 11 | 20 | 220 |

25-35 | 21 | 30 | 630 |

35-45 | 23 | 40 | 920 |

45-55 | 14 | 50 | 700 |

55-65 | 5 | 60 | 300 |

Sum f_{i} = 80 | Sum f_{i}x_{i} = 2830 |

Mean = x̄ = ∑f

_{i}x

_{i}/∑f

_{i}

= 2830/80 = 35.37 yr

2. The following data gives the information on the observed lifetimes (in hours) of 225

electrical components :

Lifetime (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |

Frequency | 10 | 35 | 52 | 61 | 38 | 29 |

Determine the modal lifetimes of the components.

**Answer**

Modal class of the given data is 60–80.

Modal class = 60-80, l = 60, f

_{m}= 61, f

_{1}= 52, f

_{2}= 38 and h = 20

3. The following data gives the distribution of total monthly household expenditure of 200

families of a village. Find the modal monthly expenditure of the families. Also, find the

mean monthly expenditure :

Expenditure | Number of families |

1000-1500 | 24 |

1500-2000 | 40 |

2000-2500 | 33 |

2500-3000 | 28 |

3000-3500 | 30 |

3500-4000 | 22 |

4000-4500 | 16 |

4500-5000 | 7 |

**Answer**

Modal class = 1500-2000, l = 1500, f

_{m}= 40, f

_{1}= 24, f

_{2}= 33 and h = 500

Calculation for mean:

Class Interval | fi | xi | di = xi - a | ui = di/h | fiui |

1000-1500 | 24 | 1250 | -1500 | -3 | -72 |

1500-2000 | 40 | 1750 | -1000 | -2 | -80 |

2000-2500 | 33 | 2250 | -500 | -1 | -33 |

2500-3000 | 28 | 2750 | 0 | 0 | 0 |

3000-3500 | 30 | 3250 | 500 | 1 | 30 |

3500-4000 | 22 | 3750 | 1000 | 2 | 44 |

4000-4500 | 16 | 4250 | 1500 | 3 | 48 |

4500-5000 | 7 | 4750 | 2000 | 4 | 28 |

fi = 200 | fiui = -35 |

Mean = x̄ = a + (∑f

_{i}u

_{i}/∑f

_{i}) х h

= 2750 + (35/200) х 500

= 2750 - 87.50 = 2662.50