**A survey was conducted by a group of students as a part of their environment awareness program, 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 | 0-1 | 2-4 | 4-6 | 6-10 | 8-10 | 10-12 | 12-14 |

number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |

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

Solution:

To find the class mark (xi) for each interval, the following relation is used.

Class mark 𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕+}^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

xi and fixi can be calculated as follows.

Number of plants | number of houses | x_{i} | f_{i}x_{i} |

0 – 2 | 1 | 1 | 1 |

2 – 4 | 2 | 3 | 2 x 3 = 6 |

4 – 6 | 1 | 5 | 1 x 5 = 5 |

6 – 8 | 5 | 7 | 5 x 7 = 35 |

8 – 10 | 6 | 9 | 6 x 9 = 54 |

10 – 12 | 2 | 11 | 2 x 11 = 22 |

12 – 14 | 3 | 13 | 3 x13 = 39 |

Total | 20 | | 162 |

From the table, it can be observed that

∑f_{i} = 20

∑f_{i}x_{i} = 162

_{= }^{162}/_{20 = 8.1}

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (xi) and fi are small.

**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.**

Solution:

To find the class mark for each interval, the following relation is used.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Class size (h) of this data = 20

Taking 150 as assured mean (a), di, ui, and fiui can be calculated as follows.

Daily wages(in Rs) | Number of workers(f_{i}) | x_{i} | d_{i} = x_{i} – 150 | u_{i}=^{di}/_{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 | 50 | -12 |

From the table, it can be observed that

∑f_{i} = 50

∑f_{i}x_{i} = -12

_{=150+(}^{-12}/_{50)20}

_{ = 150 – }^{24}/_{5}

= 150 – 4.8 = 145.2

Therefore, mean number of plants per house is 145.20

**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 allowance(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 |

Solution:

To find the class mark (xi) for each interval, the following relation is used.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Given that, mean pocket allowance,

Taking 18 as assured mean (a), d_{i} and f_{i}d_{i} are calculated as follows.

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

11 – 13 | 7 | 12 | -6 | – 42 | -24 |

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

15 – 17 | 9 | 16 | -2 | -18 | 0 |

17 – 19 | 13 | 18 | 0 | 0 | 6 |

19 – 21 | f | 20 | 2 | 2f | 20 |

21 – 23 | 5 | 22 | 4 | 20 | -12 |

23 – 25 | 4 | 24 | 6 | 24 | 24 |

total | ∑f = 44 + _{i}f
| 2f – 40 |

From above,

∑*f _{i} *= 44 +

*f*

∑*f _{i}u_{i}* = 2

*f – 40*

18 = 18 +^{(2f-40)}/_{(44+f)}

2*f* – 40 = 0

2f = 40

f = 20

Hence, the missing frequency, f, is 20.

**4: Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine 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 women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |

Solution:

To find the class mark of each interval (xi), the following relation is used.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Class size, h, of this data = 3

Taking 75.5 as assumed mean (a), di, ui, fiui are calculated as follows.

Number of heart beats per minute | Number of women fi | xi | di = xi − 75 | 𝒖_{𝒊} = ^{𝒅𝒊}/_{𝟑} | fiui |

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

∑*f _{i}* = 30

∑*f _{i}u_{i}* = 4

= 75.5+(^{4}/_{30})x3

= 75.5 + ^{4}/_{30} x 3

= 75.5 – 0.4

= 75.9

Therefore, mean hear beats per minute for these women are 75.9 beats per minute.

**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?**

Solution:

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

50 – 52 | 15 |

53 – 55 | 110 |

56 – 58 | 135 |

59 – 61 | 115 |

62 – 64 | 25 |

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 the upper class limit and 1/2 has to be subtracted from the lower class limit of each interval.

Class mark (xi) can be obtained by using the following relation.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Class size (h) of this data = 3

*Here we are using Step deviation method*

Taking 57 as assumed mean (a), di, ui, fiui are calculated as follows.

class interval | f_{i} | x_{i} | d_{i} = x_{i}-57 | u_{i} = ^{di}/_{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 | 400 | 25 |

∑*f _{i} = *a + (

^{∑fiui}/

_{∑fi})xh

= 57 + (^{25}/_{400})x3

= 57 + ^{3}/_{16} = 57 + 0.1875

= 57.1875 = 57.19

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

6:** 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.**

Solution:

To find the class mark (xi) for each interval, the following relation is used.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Class size = 50

Taking 225 as assumed mean (a), di, ui, fiui are calculated as follows.

Daily expenditure(in Rs) | f_{i} | x_{i} | d_{i} = x_{i} – 225 | u_{i} == ^{di}/_{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 | 25 | -7 |

∑*f _{i} = 25*

∑*f _{i}u_{i} = a + (*

^{∑f}

_{i}^{u}_{i}/_{∑fi})

*xh*

*= 225 + ( ^{-7}/_{25})x50*

*= 225 – 14*

*= 211*

Therefore, mean daily expenditure on food is Rs 211.

**To find out the concentration of SO2 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} | Frequency |

00.0 – 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 |

**Find the mean concentration of SO2 in the air.**

Solution:

To find the class marks for each interval, the following relation is used.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), di, ui, fiui are calculated as follows.

Concentration of SO_{2} | frequency fi | class mark xi | di = xi – 0.14 | ui = ^{di}/_{0.04} | fiui |

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 |

total | 30 | | | | -31 |

∑f_{i} = 30

∑fiui = -31

=0.14 +(^{-31}/_{30})(0.04)

=0.14 – 0.04133

=0.09867

=0.099ppm

Therefore, mean concentration of SO_{2} in the air is 0.099 ppm.

**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 | 38-49 |

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

*Solution:*

To find the class mark of each interval, the following relation is used.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Taking 17 as assumed mean (a), di and fidi are calculated as follows.

Number of days | number of students fi | xi | di = xi – 17 | fidi |

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 |

*∑fi = 40*

*∑fidi = -181*

*= 17 + *(^{-181}/_{40})

= 17 – 4.525

= 12.475

= 12.48

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

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 | 85-95 |

number of cities | 3 | 10 | 11 | 8 | 3 |

Solution:

To find the class marks, the following relation is used.

𝒙𝒊 = ^{𝑼𝒑𝒑𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕 + }^{𝑳𝒐𝒘𝒆𝒓 }^{𝒍𝒊𝒎𝒊𝒕}/_{𝟐}

Class size (h) for this data = 10 Taking 70 as assumed mean (a), di, ui, and fiui are calculated as follows.

Literacy rate(in%) | number of cities fi | xi | di = xi – 17 | ui = ^{di}/_{10} | fidi |

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 |

*∑fi = 35*

*∑fidi = -2*

*= 70 + *(^{-2}/_{35})x10

= 17 – ^{20}/_{35 }= 17 – ^{4}/_{7}

= 70 – 0.57

= 69.43

Therefore, mean literacy rate is 69.43%

## One thought on “Statistics- Exercise 14.1 – Class 10”

Comments are closed.