Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.2
Question 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.
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 =
On substituting the values in the formula, we get
Mode =
= 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 Interval Frequency (fi) Mid-point (xi) fixi 5-15 6 10 60 15-25 11 20 220 25-35 21 30 630 35-45 23 40 920 45-55 14 50 700 55-65 5 60 300 Sum fi = 80 Sum fixi = 2830 Mean =
= ∑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) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
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 =
On substituting the values in the formula, we get
Mode =
=
= 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:
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 |
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 =
On substituting the values in the formula, we get
Mode =
=
= 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 Interval fi xi di = xi – a ui = di/h fiui 1000-1500 24 1250 -1500 -3 -72 1500-2000 40 1750 -1000 -2 -80 2000-2500 33 2250 -500 -1 -33 2500-3000 28 2750 0 0 0 3000-3500 30 3250 500 1 30 3500-4000 22 3750 1000 2 44 4000-4500 16 4250 1500 3 48 4500-5000 7 4750 2000 4 28 fi = 200 fiui = -35 Mean =
On substituting the values in the given formula
=
= 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 teacher | Number of states / U.T |
15-20 | 3 |
20-25 | 8 |
25-30 | 9 |
30-35 | 10 |
35-40 | 3 |
40-45 | 0 |
45-50 | 0 |
50-55 | 2 |
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 =
On substituting the values in the formula, we get
Mode =
= 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 Interval Frequency (fi) Mid-point (xi) fixi 15-20 3 17.5 52.5 20-25 8 22.5 180.0 25-30 9 27.5 247.5 30-35 10 32.5 325.0 35-40 3 37.5 112.5 40-45 0 42.5 0 45-50 0 47.5 0 50-55 2 52.5 105.5 Sum fi = 35 Sum fixi = 1022.5 Mean =
= 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 Scored | Number of Batsman |
3000-4000 | 4 |
4000-5000 | 18 |
5000-6000 | 9 |
6000-7000 | 7 |
7000-8000 | 6 |
8000-9000 | 3 |
9000-10000 | 1 |
10000-11000 | 1 |
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 =
On substituting the values in the formula, we get
Mode =
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 cars | Frequency |
0-10 | 7 |
10-20 | 14 |
20-30 | 13 |
30-40 | 12 |
40-50 | 20 |
50-60 | 11 |
60-70 | 15 |
70-80 | 8 |
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 =
On substituting the values in the formula, we get
Mode =
Mode = 40 + 80/17 = 40 + 4.7 = 44.7
Hence, the mode of the given data is 44.7 cars
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