TOPICS

Exercise - 14.1

Statistics

**Question-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
Which method did you use for finding the mean, and why?

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2

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

Total | Σf_{i} = 20 | Σf_{i}x_{i} = 162 |

**Question-2 :-** Consider the following distribution of daily wages of 50 workers of a factory
Find the mean daily wages of the workers of the factory by using an appropriate method

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 20 Taking assumed mean (a) = 150

Daily wages (in Rs.) | Number of workers (f_{i}) | x_{i} | d_{i} = x_{i} - 150 | u_{i} = d_{i}/20 | 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 | Σf_{i} = 50 | Σf_{i}u_{i} = -12 |

**Question-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.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Given that the mean pocket allowance x = Rs. 18 Taking assumed mean (a) = 18

Daily pocket allowance (in Rs.) | Number of children (f_{i}) | 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 | Σf_{i} = 44 + f | Σf_{i}d_{i} = 2f - 40 |

**Question-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.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 3 Taking assumed mean (a) = 75.5

Number of heart beats (per min.) | Number of women (f_{i}) | x_{i} | d_{i} = x_{i} - 75.5 | u_{i} = d_{i}/3 | 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 | Σf_{i} = 30 | Σf_{i}u_{i} = 4 |

**Question-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?

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, 1/2 has to be added to upper class limit and 1/2 has to be substracted from the lower class limit of each interval. From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 3 Taking assumed mean (a) = 57

Number of mangoes | Number of boxes (f_{i}) | x_{i} | d_{i} = x_{i} - 57 | u_{i} = d_{i}/3 | 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 | Σf_{i} = 400 | Σf_{i}u_{i} = 25 |

**Question-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.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 50 Taking assumed mean (a) = 225

Daily expenditure (in Rs.) | Number of households (f_{i}) | x_{i} | d_{i} = x_{i} - 225 | u_{i} = d_{i}/50 | 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 | Σf_{i} = 25 | Σf_{i}u_{i} = -7 |

**Question-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:
Find the mean concentration of SO_{2} in the air.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 0.04 Taking assumed mean (a) = 0.14

Concentration of SO_{2} (in ppm) | Frequency (f_{i}) | x_{i} | d_{i} = x_{i} - 0.14 | u_{i} = d_{i}/0.04 | 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 | 0.4 | 1 | 4 |

0.20 - 0.24 | 2 | 0.22 | 0.8 | 2 | 4 |

Total | Σf_{i} = 30 | Σf_{i}u_{i} = -31 |

**Question-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.

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Taking assumed mean (a) = 17

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 | Σf_{i} = 40 | Σf_{i}d_{i} = -181 |

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

From given table, to find the class mark (x_{i}) for each inerval, the following relation is used. Class mark (x_{i}) = (upper class limit + lower class limit)/2 Class size (h) = 10 Taking assumed mean (a) = 70

Literacy rate (in %) | Number of cities (f_{i}) | x_{i} | d_{i} = x_{i} - 70 | u_{i} = d_{i}/10 | 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 | Σf_{i} = 35 | Σf_{i}u_{i} = -2 |

CLASSES