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Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.2

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  • Last Updated : 03 Mar, 2021
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Question 1. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)5-1515-2525-3535-4545-5555-65
Number of patients6112123145

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

Solution:

The greatest frequency in the given table is 23, so the modal class = 35 – 45,

l = 35,

Class width = 10, and the frequencies are

fm = 23, f1 = 21 and f2 = 14

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 35+\left[\frac{(23-21)}{(46-21-14)}\right]×10

= 35 + (20/11) = 35 + 1.8

= 36.8

Hence, the mode of the given data is 36.8 year

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Class IntervalFrequency (fi)Mid-point (xi)fixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
 Sum fi = 80 Sum fixi = 2830

Mean = \bar{x} = ∑fixi /∑fi

= 2830/80

= 35.37 years

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

Lifetime (in hours)0-2020-4040-6060-8080-100100-120
Frequency103552613829

Determine the modal lifetimes of the components.

Solution:

According to the given question

The modal class is 60 – 80

l = 60, and the frequencies are

fm = 61, f1 = 52, f2 = 38 and h = 20

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 60+\left[\frac{(61-52)}{(122-52-38)}\right]×20

60+\frac{(9 \times 20)}{32}

= 60 + 45/8 = 60 + 5.625

Hence, the modal lifetime of the components is 65.625 hours.

Question 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:

ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007

Solution:

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

fm = 40 f1 = 24, f2 = 33 and

h = 500

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 1500+\left[\frac{(40-24)}{(80-24-33)}\right]×500

1500+\frac{16×500}{23}

= 1500 + 8000/23 = 1500 + 347.83

So, the modal monthly expenditure of the families is 1847.83 Rupees 

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Let us considered a mean, A be 2750

Class Intervalfixidi = xi – aui = di/hfiui
1000-1500241250-1500-3-72
1500-2000401750-1000-2-80
2000-2500332250-500-1-33
2500-3000282750000
3000-3500303250500130
3500-40002237501000244
4000-45001642501500348
4500-5000747502000428
 fi = 200   fiui = -35

Mean = \overline{x} = a +\frac{∑f_iu_i}{∑f_i}×h

On substituting the values in the given formula

2750+\frac{-35}{200}×500

= 2750 – 87.50

= 2662.50

Hence, the mean monthly expenditure of the families is  2662.50 Rupees

Question 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

No of Students per teacherNumber of states / U.T
15-203
20-258
25-309
30-3510
35-403
40-450
45-500
50-552

Solution:

According to the question

Modal class = 30 – 35,

l = 30,

Class width (h) = 5, and the frequencies are

fm = 10, f1 = 9 and f2 = 3

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 30+\frac{(10-9)}{(20-9-3)}×5

= 30 + 5/8 = 30 + 0.625

= 30.625

Hence, the mode of the given data is 30.625

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Class IntervalFrequency (fi)Mid-point (xi)fixi
15-20317.552.5
20-25822.5180.0
25-30927.5247.5
30-351032.5325.0
35-40337.5112.5
40-45042.50
45-50047.50
50-55252.5105.5
 Sum fi = 35 Sum fixi = 1022.5

Mean = \bar{x} = \frac{∑f_ix_i }{∑f_i}

= 1022.5/35 

= 29.2

Hence, the mean is 29.2

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

Run ScoredNumber of Batsman
3000-40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-100001
10000-110001

Find the mode of the data.

Solution:

According to the question

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000, and the frequencies are

fm = 18, f1 = 4 and f2 = 9

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 4000+\frac{(18-4)}{(36-4-9)}×1000

Mode = 4000 + 14000/23 = 4000 + 608.695

= 4608.695

Hence, the mode of the given data is 4608.7 runs

Question 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:

Number of carsFrequency
0-107
10-2014
20-3013
30-4012
40-5020
50-6011
60-7015
70-808

Solution:

According to the question

Modal class = 40 – 50, l = 40,

Class width (h) = 10, and the frequencies are

fm = 20, f1 = 12 and f2 = 11

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 40+\frac{(20-12)}{(40-12-11)}×10

Mode = 40 + 80/17 = 40 + 4.7 = 44.7

Hence, the mode of the given data is 44.7 cars


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