EXERCISE 14.1

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.

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?

Answer:

No. of plants (Class interval)	No. of houses (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

Mean = x̄ = ∑f_ix_i /∑f_i = 162/20 = 8.1

We would use direct method because the numerical value of f_i and x_i are small.

2. 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.

Answer:
Here, the value of mid-point (x_i) is very large, so assumed mean A = 150 and class interval is h = 20.
So, u_i = (x_i – A)/h = u_i = (x_i – 150)/20

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

Mean = x̄ = A + h∑f_iu_i /∑f_i =150 + (20 × -12/50) = 150 – 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20

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.

Answer:

Here, the value of mid-point (x_i)  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

Mean = x̄ = ∑f_ix_i /∑f_i = (752+20f)/(44+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

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:

Answer:

x_i = (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5

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∑f_iu_i /∑f_i = 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number 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?

Answer:
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3

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

Mean = x̄ = A + ∑f_id_i /∑f_i = 57 + (75/400) = 57 + 0.1875 = 57.19

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.

Answer:

Find the midpoint of the given interval using the formula.

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

Let is assume the mean (A) = 225

Class size (h) = 50

= 225+50(-7/25)

= 225-14

= 211

Therefore, the mean daily expenditure on food is 211

7. 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:

Find the mean concentration of SO₂ in the air.

Answer:

The formula to find out the mean is

Mean = x̄ = ∑f_ix_i /∑f_i

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO₂ in air is 0.099 ppm.

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.

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:

Find the midpoint of the given interval using the formula.

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

The mean formula is,

Mean = x̄ = ∑f_ix_i /∑f_i

= 499/40

= 12.48 days

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

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	85-98
Number of cities	3	10	11	8	3

Answer:

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

Mean = x̄ = a + (∑f_iu_i /∑f_i) х h
= 70 + (-2/35) х 10 = 69.42

EXERCISE 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency.

Answer:

To find out the modal class, let us the consider the class interval with high frequency

Here, the greatest frequency = 23, so the modal class = 35 – 45,

l = 35,

class width (h) = 10,

f_m = 23,

f₁ = 21 and f₂ = 14

The formula to find the mode is

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

So the mode of the given data = 36.8 year

Calculation of Mean:

First find the midpoint using the formula, x_i = (upper limit +lower limit)/2

The mean formula is

Mean = x̄ = ∑f_ix_i /∑f_i

= 2830/80

= 35.37 years

Therefore, the mean of the given data = 35.37 years.

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Lifetime (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Answer:

Given:

From the given data the modal class is 60–80.

l = 60,

The frequencies are:

f_m = 61, f₁ = 52, f₂ = 38 and h = 20

The formula to find the mode is

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 20)/32)

Mode = 60+(45/8) = 60+ 5.625

Therefore, modal lifetime of the components = 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Expenditure	Number of families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

Answer:

For Mode:

Given data:

Modal class = 1500-2000,

l = 1500,

Frequencies:

f_m = 40 f₁ = 24, f₂ = 33 and

h = 500

Mode formula:

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

Mode =1500+[(40-24)/(80-24-33)]×500

Mode = 1500+((16×500)/23)

Mode = 1500+(8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, x_i =(upper limit +lower limit)/2

Let us assume a mean, A be 2750

The formula to calculate the mean,

Mean = x̄ = a +(∑f_iu_i /∑f_i)×h

Substitute the values in the given formula

= 2750+(-35/200)×500

= 2750-87.50

= 2662.50

So, the mean monthly expenditure of the families = Rupees 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Answer:

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day cricket matches:

Find mode of the data.

Answer:

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below:

Find the mode of the data.

Answer:

EXERCISE 14.3

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Answer:

 

2. If the median of the distribution given below is 28.5, then find the values of x and y.

Answer:

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.

Answer:

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter and data obtained is represented in the following table. Find the median length of the leaves.

Answer:

EXERCISE 14.4

1. The following distribution gives the daily income of 50 workers of a factory:

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Answer:

Now, by drawing the points on the graph,

i.e., (120, 12); (140, 26); 160, 34); (180, 40); (200, 50)

2. During the medical checkup of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Answer:

Hence, the points for graph are:

(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution and draw its ogive.

Answer:

The points for the graph are:

(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16)