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
1** :

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 |

**Solution:**

In order to find themean value, we will use direct method because the numerical value of f_{i} and x_{i} are small.

Find the midpoint ofthe given interval using the formula.

Midpoint (x_{i}) = (upper limit + lower limit)/2

No. of plants (Class interval) | No. of houses Frequency (f | Mid-point (x | f |

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 | Sum f |

The formula to findthe mean is:

Mean = x̄ = ∑f_{i }x_{i} /∑f_{i }

= 162/20

= 8.1_{}

Therefore, the mean number of plants per house is 8.1

Consider the following distribution of daily wages of 50 workers of afactory.

**Answer
2** :

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 anappropriate method.

Solution:

Find the midpoint ofthe given interval using the formula.

Midpoint (x_{i}) = (upper limit + lower limit)/2

In this case, thevalue of mid-point (x_{i}) is very large, so let us assume the meanvalue, A = 150 and class interval is h = 20.

So, u_{i }= (x_{i} – A)/h = u_{i }= (x_{i} – 150)/20

Substitute and findthe values as follows:

Daily wages (Class interval) | Number of workers frequency (f | Mid-point (x | u | f |

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 | Sum f |

So, the formula tofind out the mean is:

Mean = x̄ = A + h∑f_{i}u_{i} /∑f_{i }=150+ (20 × -12/50) = 150 – 4.8 = 145.20

Thus, mean daily wageof the workers = Rs. 145.20

The following distribution shows the daily pocket allowance of childrenof a locality. The mean pocket allowance is Rs 18. Find the missing frequencyf.

**Answer
3** :

Daily Pocket Allowance(in c) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-35 |

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

Solution:

To find out themissing frequency, use the mean formula.

Here, the value ofmid-point (x_{i}) mean x̄ = 18

Class interval | Number of children (f | Mid-point (x | f |

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 | Sum f |

The mean formula is

Mean = x̄ = ∑f_{i}x_{i} /∑f_{i }=(752+20f)/(44+f)

Now substitute thevalues and equate to find the missing frequency (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

So, the missingfrequency, f = 20.

**Answer
4** :

Find the mean heart beats per minutefor 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:

From the given data, let us assume the mean asA = 75.5

x_{i }= (Upper limit + Lowerlimit)/2

Class size (h) = 3

Now, find the u_{i }and f_{i}u_{i} asfollows:

Class Interval | Number of women (f | Mid-point (x | u | f |

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 | Sum f |

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

Therefore, the mean heart beats per minute forthese women is 75.9

**Answer
5** :

The following was the distribution of mangoesaccording 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 numberof mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

Since, the given data is not continuous so weadd 0.5 to the upper limit and subtract 0.45 from the lower limit as the gapbetween two intervals are 1

Here, assumed mean (A) = 57

Class size (h) = 3

Here, the step deviation is used because thefrequency values are big.

Class Interval | Number of boxes (f | Mid-point (x | d | f |

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 | Sum f |

The formula to find out the Mean is:

Mean = x̄ = A +h ∑f_{i}d_{i} /∑f_{i }

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept ina packing box is 57.19

**Answer
6** :

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

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

Solution:

Find the midpoint of the given interval usingthe formula.

Midpoint (x_{i}) = (upper limit +lower limit)/2

Let is assume the mean (A) = 225

Class size (h) = 50

Class Interval | Number of households (f | Mid-point (x | d | u | f |

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 |

Sum f | Sum f |

Mean = x̄ = A +h∑f_{i}u_{i} /∑f_{i}

_{ }=225+50(-7/25)

= 225-14

= 211

Therefore, the mean daily expenditure on foodis 211

**Answer
7** :

Concentration of SO | 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 |

Find the meanconcentration of SO_{2} in the air.

Solution:

To find out the mean, first find the midpointof the given frequencies as follows:

Concentration of SO | Frequency (f | Mid-point (x | f |

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 | Sum (f |

The formula to find out the mean is

Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}

= 2.96/30

= 0.099 ppm

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

A class teacher hasthe following absentee record of 40 students of a class for the whole

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

**Answer
8** :

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 |

Solution:

Find the midpoint of the given interval usingthe formula.

Midpoint (x_{i}) = (upper limit +lower limit)/2

Class interval | Frequency (f | Mid-point (x | f |

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 | Sum f |

The mean formula is,

Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}

= 499/40

= 12.48 days

Therefore, the mean number of days a studentwas absent = 12.48.

literacy rate.

**Answer
9** :

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

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

Solution:

Find the midpoint of the given interval usingthe formula.

Midpoint (x_{i}) = (upper limit +lower limit)/2

In this case, the value of mid-point (x_{i})is very large, so let us assume the mean value, A = 70 and class interval is h= 10.

So, u_{i }= (x_{i}-A)/h =u_{i }= (x_{i}-70)/10

Substitute and find the values as follows:

Class Interval | Frequency (f | (x | d | u | f |

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 | Sum f |

So, Mean = x̄ = A+(∑f_{i}u_{i} /∑f_{i})×h

= 70+(-2/35)×10

= 69.42

Therefore, the mean literacy part = 69.42

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