## Statistics

###
*(NCERT Ex 14.1)
*

*(NCERT Ex 14.1)*

**Q1: 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.**

Answer:

We use the following relation to find the class mark (x

_{i}) for each interval:

Class mark (x

_{i}) =

Calculating x

_{i}and f

_{i}x

_{i}as follows:

Number of plants | Number of houses(f_{i}) | x_{i} | f_{i}x_{i} |

0 -2 | 1 | 1 | 1 ☓ 1 = 1 |

2 -4 | 2 | 3 | 2 ☓ 3 = 6 |

4 -6 | 1 | 5 | 1 ☓ 5 = 5 |

6 -8 | 5 | 7 | 5 ☓ 7 = 35 |

8 - 10 | 6 | 9 | 6 ☓ 9 = 54 |

10 -12 | 2 | 11 | 2 ☓ 11 = 22 |

12 - 14 | 3 | 13 | 3 ☓ 13 = 39 |

Total | 20 | 162 |

From the table, it is observed that

Σ f

_{i}= 20

Σ f

_{i}x

_{i}= 162

Mean , x =

= = 8.1

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (x

_{i}) and f

_{i}are small.

**Q2: Consider the following distribution of daily wages of 50 worker 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 |

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer:

To find the class mark (x

_{i}) for each interval, using the following relation:

Class mark (x

_{i}) =

Class size (h) = 20

Taking 150 as assumed mean (a), d

_{i}, u

_{i}, and f

_{i}u

_{i}can be calculated as follows:

Daily wages(in Rs) | Number of workers(f_{i}) | x_{i} | d_{i} = x_{i} - 150 | u = _{i} | f_{i}u_{i} |

100 - 120 | 12 | 110 | -40 | -2 | -24 |

120 - 140 | 14 | 130 | -20 | -1 | -14 |

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

160 - 180 | 6 | 170 | 20 | 1 | 6 |

180 - 200 | 10 | 190 | 40 | 2 | 20 |

Total | 50 | -12 |

Σ f

_{i}= 50

Σ f

_{i}u

_{i}= -12

Mean , x = a +()h

= 150 + ()20

= 150 -

= 150 - 4.8

= 145.2

∴ the mean daily wage of the workers of the factory =

**Rs 145.20**.

**Q3: 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.**

Daily pocket allownace(in Rs) | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |

Number of workers | 7 | 6 | 9 | 13 | f | 5 | 4 |

Answer:

To find the class mark (x

_{i}) for each interval, using the following relation:

x

_{i}=

Given that, mean pocket allowance, x = Rs 18

Taking 18 as assumed mean (a), d

_{i}and f

_{i}d

_{i}are calculated as:

Daily pocket allowance(in Rs) | Number of children(f_{i}) | Class mark(x_{i}) | d_{i} = x_{i} - 18 | f_{i}d_{i} |

11 - 13 | 7 | 12 | -6 | -42 |

13 - 15 | 6 | 14 | -4 | -24 |

15 - 17 | 9 | 16 | -2 | -18 |

17 - 19 | 13 | 18 | 0 | 0 |

19 - 21 | f | 20 | 2 | 2f |

21 - 23 | 5 | 22 | 4 | 20 |

23 - 25 | 4 | 24 | 6 | 24 |

Total | Σ f_{i} = 44 + f | 2f - 40 |

From the table:

Σ f

_{i}= 44 + f

Σ f

_{i}d

_{i}= 2f - 40

x = a +

18 = 18 + ()

0 =

2f - 40 = 0

2f = 40

f = 20

∴ the missing frequency, f, is 20.

**Q4: Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.**

Number of heart beats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |

Number of woman | 2 | 4 | 3 | 8 | 7 | 4 | 2 |

Answer:

To find the class mark (x

_{i}) for each interval, using the following relation:

x

_{i}=

Class size, h, of this data = 3

Considering 75.5 as assumed mean (a), d

_{i}, u

_{i}, f

_{i}u

_{i}are calculated as follows.

Number of heart beats per minute | Number of woman(f_{i}) | x_{i} | d_{i} = x_{i} - 75.5 | u = _{i} | f_{i}u_{i} |

65 - 68 | 2 | 66.5 | -9 | -3 | -6 |

68 - 71 | 4 | 69.5 | -6 | -2 | -8 |

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

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

77 - 80 | 7 | 78.5 | 3 | 1 | 7 |

80 - 83 | 4 | 81.5 | 6 | 2 | 8 |

83 - 86 | 2 | 84.5 | 9 | 3 | 6 |

Total | 30 | 4 |

From the table:

Σ f

_{i}= 30

Σ f

_{i}u

_{i}= 4

Mean , x = a +()☓ h

= 75.5 + 4/30 ☓ 3

= 75.5 + 0.4 = 75.9

∴ mean heart beats per minute for these women are 75.9 beats per minute.

**Q5: 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.**

Number of mangoes | 50 - 52 | 53 - 55 | 56 - 58 | 59 - 61 | 62 - 64 |

Number of boxes | 15 | 110 | 135 | 115 | 25 |

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

Answer:

Number of mangoes | Number of boxes f_{i} |

50 - 52 | 15 |

53 - 55 | 110 |

56 - 58 | 135 |

59 - 61 | 115 |

62 - 64 | 25 |

Since the class intervals are not continuous, there is a gap of 1 between two class intervals. Therefore adding to the upper class limit and subtract from the lower class limit of each interval.

Obtaining Class mark (x

_{i}) by using the following relation.

x

_{i}=

Class size (h) of this data = 3

Taking 57 as assumed mean (a), d

_{i}, u

_{i}, f

_{i}u

_{i }are calculated as:

Class interval | f_{i} | x_{i} | d_{i} = x_{i} - 57 | u = _{i} | f_{i}u_{i} |

49.5 - 52.5 | 15 | 51 | -6 | -2 | -30 |

52.5 - 55.5 | 110 | 54 | -3 | -1 | -110 |

55.5 - 58.5 | 135 | 57 | 0 | 0 | 0 |

58.5 - 61.5 | 115 | 60 | 3 | 1 | 115 |

61.5 - 64.5 | 25 | 63 | 6 | 2 | 50 |

Total | 400 | 25 |

From the table:

Σ f

_{i}= 400

Σ f

_{i}u

_{i}= 25

Mean , x = a +()☓ h

= 57 + ()☓ 3

= 57 + 3/16= 57 + 0.1875

= 57.1875

≈ 57.19

Mean number of mangoes kept in a packing box = 57.19.

Using

*step deviation*method as the values of f

_{i}, d

_{i}are big and also, there is a common multiple between all d

_{i}.

**Q6: The table below shows the daily expenditure on food of 25 households in a locality.**

Daily expenditure(in Rs) | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |

Number of households | 4 | 5 | 12 | 2 | 2 |

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

Answer:

To find the class mark (x

_{i}) for each interval, using the following relation:

x

_{i}=

Class size = 50

Taking 225 as assumed mean (a), d

_{i}, u

_{i}, f

_{i}u

_{i}are calculated as:

Daily expenditure(in Rs) | f_{i} | x_{i} | d_{i} = x_{i} - 225 | u = _{i} | f_{i}u_{i} |

100 - 150 | 4 | 125 | -100 | -2 | -8 |

150 - 200 | 5 | 175 | -50 | -1 | -5 |

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

250 - 300 | 2 | 275 | 50 | 1 | 2 |

300 - 350 | 2 | 325 | 100 | 2 | 4 |

Total | 25 | -7 |

From the table:

Σ f

_{i}= 25

Σ f

_{i}u

_{i}= -7

Mean , x = a +()☓ h

= 225 + ()☓ 50

= 225 - 14

= 211

Thus the mean daily expenditure on food is Rs 211.

**Q7: 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:concentration of SO_{2} (in ppm) | Frequency |

0.00 - 0.04 | 4 |

0.04 - 0.08 | 9 |

0.08 - 0.12 | 9 |

0.12 - 0.16 | 2 |

0.16 - 0.20 | 4 |

0.20 - 0.24 | 2 |

Answer:

To find the class mark (x

_{i}) for each interval, using the following relation:

x

_{i}=

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), calculating d

_{i}, u

_{i}, f

_{i}u

_{i}as:

concentration of SO_{2} (in ppm) | Frequency(f_{i}) | Class mark(x_{i}) | d_{i} = x_{i} - 0.14 | u = _{i} | f_{i}u_{i} |

0.00 - 0.04 | 4 | 0.02 | -0.12 | -3 | -12 |

0.04 - 0.08 | 9 | 0.06 | -0.08 | -2 | -18 |

0.08 - 0.12 | 9 | 0.10 | -0.04 | -1 | -9 |

0.12 - 0.16 | 2 | 0.14 | 0 | 0 | 0 |

0.16 - 0.20 | 4 | 0.18 | 0.04 | 1 | 4 |

0.20 - 0.24 | 2 | 0.22 | 0.08 | 2 | 4 |

From the table:

Σ f

_{i}= 30

Σ f

_{i}u

_{i}= -31

Mean , x = a +()☓ h

= 0.14 + ()☓ 0.04

= 0.14 - 0.04133

= 0.09867

≈ 0.099 ppm

Therefore, mean concentration of SO

_{2}in the air =

**0.099 ppm**.

**Q8: 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:

To find the class mark (x

_{i}) for each interval, using the following relation:

x

_{i}=

Taking 17 as assumed mean (a), calculating d

_{i}and f

_{i}d

_{i}as:

Number of days | Number of students(f_{i}) | x_{i} | d_{i} = x_{i} - 17 | f_{i}d_{i} |

0 - 6 | 11 | 3 | - 14 | - 154 |

6 - 10 | 10 | 8 | - 9 | - 90 |

10 - 14 | 7 | 12 | - 5 | - 35 |

14 - 20 | 4 | 17 | 0 | 0 |

20 - 28 | 4 | 24 | 7 | 28 |

28 - 38 | 3 | 33 | 16 | 48 |

38 - 40 | 1 | 39 | 22 | 22 |

Total | 40 | - 181 |

From the table:

Σ f

_{i}= 40

Σ f

_{i}d

_{i}= -181

Mean, x = a +

= 17 + ()

= 17 - 4.525

= 12.475

≈12.48

Therefore, the mean number of days = 12.48 days for which a student was absent.

**Q9: 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 - 95 |

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

Answer:

To find the class mark (x

_{i}) for each interval, using the following relation:

x

_{i}=

Class size(h) for this data = 10

Taking 70 as assumed mean (a), calculating d

_{i}, u

_{i}, f

_{i}u

_{i :}

Literacy rate(in %) | Number of cities(f_{i}) | x_{i} | d_{i} = x_{i} - 70 | u = _{i} | 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 |

Total | 35 | -2 |

From the table:

Σ f

_{i}= 35

Σ f

_{i}u

_{i}= -2

Mean , x = a +()☓ h

= 70 + ()☓ 10

= 70 -

= 70 -

= 70 - 0.57

= 69.43

Therefore, mean literacy rate is 69.43%.

Very good!

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