Class 10th, Statistics, 14.1 (Solution)

Class 10th,- Statistics

 Exercise: 14.1 (Solution), Q1 to Q9 

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Q1. 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	0-2	2-4	4-6	6-8	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: In order to find the mean value,
we will use direct method because the numerical value of fi and xi are small.
Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2

No. of plants (Class interval)	No. of houses Frequency (fi)	Mid-point (xi)	fixi
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 fi = 20		Sum fixi = 162

The formula to find the mean is: Mean = x̄ = ∑fi xi /∑fi 
= 162/20
= 8.1
Therefore, the mean number of plants per house is 8.1

Q2. 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

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution: Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large,so let us assume the mean value, A = 150 and class interval is h = 20.

So, ui = (xi – A)/h = ui = (xi – 150)/20

Substitute and find the values as follows:

Daily wages (Class interval)	Number of workers frequency (fi)	Mid-point (xi)	ui = (xi – 150)/20	fiui
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 fi = 50			Sum fiui = -12

So, the formula to find out the mean is:
Mean = x̄ = A + h∑fiui /∑fi 
=150 + (20 × -12/50)
= 150 – 4.8
= 145.20
Thus, mean daily wage of the workers = 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 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 the missing frequency, use the mean formula. Here, the value of mid-point (xi)  mean x̄ = 18

Class interval	Number of children (fi)	Mid-point (xi)	fixi
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	fi = 44+f		Sum fixi = 752+20f

The mean formula is Mean = x̄ = ∑fixi /∑fi 

= (752+20f)/(44+f)

Now substitute the values 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 missing frequency, f = 20.

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

Number of heartbeats 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 as A = 75.5
xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Now, find the ui and fi ui as follows:

Class Interval	Number of women (fi)	Mid-point (xi)	ui = (xi – 75.5)/h	fiui
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 fi= 30			Sum fiui = 4

Mean = x̄ = A + h∑fiui /∑fi 
= 75.5 + 3×(4/30) 75.5 + 4/10
= 75.5 + 0.4
= 75.9
Therefore, the mean heart beats per minute for these women is 75.9

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?
Solution: Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57 Class size (h) = 3 Here, the step deviation is used because the frequency values are big.

Class Interval	Number of boxes (fi)	Mid-point (xi)	di = xi – A	fidi
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 fi = 400			Sum fidi = 75

The formula to find out the Mean is:

Mean = x̄ = A +h ∑fidi /∑fi 

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

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

Q6. 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.

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 using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Let is assume the mean (A) = 225 Class size (h) = 50

Class Interval	Number of households (fi)	Mid-point (xi)	di = xi – A	ui=di/50	fiui
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 fi = 25				Sum fiui = -7

Mean = x̄ = A +h∑fiui /∑fi

= 225 + 50(-7/25)

= 225 – 14

= 211

Therefore, the mean daily expenditure on food is 211

Q7. 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 SO2 ( 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

Find the mean concentration of SO2 in the air.

Solution: To find out the mean, first find the midpoint of the given frequencies as follows:

Concentration of SO2 (in ppm)	Frequency (fi)	Mid-point (xi)	fixi
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 fi = 30		Sum (fixi) = 2.96

The formula to find out the mean is
Mean = x̄ = ∑fixi /∑fi
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO2 in air is 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

Solution: Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Class interval	Frequency (fi)	Mid-point (xi)	fixi
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 fi = 40		Sum fixi = 499

The mean formula is,
Mean = x̄ = ∑fixi /∑fi
= 499/40
= 12.48 days

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

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	85-98
Number of cities	3	10	11	8	3

Solution: Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large,

so let us assume the mean value, A = 70 and class interval is h = 10.

So, ui = (xi – A)/h

= ui 

= (xi – 70)/10

Substitute and find the values as follows:

Class Interval	Frequency (fi)	(xi)	di = xi – a	ui = di/h	fiui
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 fi  = 35				Sum fiui  = -2

So, Mean = x̄ = A + (∑fiui /∑fi) х h

= 70 + (-2/35) х 10

= 69.42

Therefore, the mean literacy part

= 69.42

See next blog for more Solution.      Thank you.