NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.2#
Exercise 13.2 covers finding the mode of grouped data using the mode formula.
Mode of Grouped Data:
$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$where $l$ = lower class limit of modal class, $f_1$ = frequency of modal class, $f_0$ = frequency of class preceding modal class, $f_2$ = frequency of class succeeding modal class, $h$ = class size.
1. The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. #
Answer
| Age (years) | Number of patients ($f_i$) | $x_i$ | $f_i x_i$ |
|---|---|---|---|
| 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 |
| Total | 80 | 2830 |
Mean:
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2830}{80} = 35.375$$So the mean age is about 35.37 years.
Mode: Modal class = 35-45 (highest frequency = 23)
$l = 35$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$
Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:
$$= 35 + \frac{23 - 21}{2(23) - 21 - 14} \times 10$$$$= 35 + \frac{2}{11} \times 10$$
$$= 35 + 1.82$$
$$= 36.82 \text{ (about)}$$
So the mode is about 36.8 years.
∴ Mode ≈ 36.8 years and mean ≈ 35.37 years.
Interpretation: The largest number of patients admitted are about 36.8 years old, while the average age of a patient admitted is about 35.37 years.
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components. #
Answer
| Lifetime (hours) | Frequency |
|---|---|
| 0-20 | 10 |
| 20-40 | 35 |
| 40-60 | 52 |
| 60-80 | 61 |
| 80-100 | 38 |
| 100-120 | 29 |
Modal class = 60-80 (highest frequency = 61)
$l = 60$, $f_1 = 61$, $f_0 = 52$, $f_2 = 38$, $h = 20$
Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:
$$= 60 + \frac{61 - 52}{2(61) - 52 - 38} \times 20$$$$= 60 + \frac{9}{32} \times 20$$
$$= 60 + 5.625$$
$$= 65.625$$
∴ The modal lifetime of the components is about 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. #
Answer
| Expenditure (₹) | $f_i$ | $x_i$ | $f_i x_i$ |
|---|---|---|---|
| 1000-1500 | 24 | 1250 | 30000 |
| 1500-2000 | 40 | 1750 | 70000 |
| 2000-2500 | 33 | 2250 | 74250 |
| 2500-3000 | 28 | 2750 | 77000 |
| 3000-3500 | 30 | 3250 | 97500 |
| 3500-4000 | 22 | 3750 | 82500 |
| 4000-4500 | 16 | 4250 | 68000 |
| 4500-5000 | 7 | 4750 | 33250 |
| Total | 200 | 532500 |
Mean:
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{532500}{200} = 2662.50$$Mode: Modal class = 1500-2000 ($f_1 = 40$ is highest)
$l = 1500$, $f_1 = 40$, $f_0 = 24$, $f_2 = 33$, $h = 500$
Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:
$$= 1500 + \frac{40 - 24}{2(40) - 24 - 33} \times 500$$$$= 1500 + \frac{16}{23} \times 500$$
$$= 1500 + 347.83$$
$$= 1847.83 \text{ (about)}$$
∴ Modal expenditure ≈ ₹1847.83, Mean expenditure = ₹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
| Number of students per teacher | $f_i$ | $x_i$ | $f_i x_i$ |
|---|---|---|---|
| 15-20 | 3 | 17.5 | 52.5 |
| 20-25 | 8 | 22.5 | 180 |
| 25-30 | 9 | 27.5 | 247.5 |
| 30-35 | 10 | 32.5 | 325 |
| 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 |
| Total | 35 | 1022.5 |
Mean:
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1022.5}{35} = 29.2 \text{ (about)}$$Mode: Modal class = 30-35 ($f_1 = 10$)
$l = 30$, $f_1 = 10$, $f_0 = 9$, $f_2 = 3$, $h = 5$
Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:
$$= 30 + \frac{10 - 9}{2(10) - 9 - 3} \times 5$$$$= 30 + \frac{1}{8} \times 5$$
$$= 30.625$$
∴ Mode ≈ 30.6 and mean ≈ 29.2.
Interpretation: Most states/U.T. have a teacher-student ratio of about 30.6, and the average ratio is about 29.2.
5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data. #
Answer
| Runs scored | $f_i$ |
|---|---|
| 3000-4000 | 4 |
| 4000-5000 | 18 |
| 5000-6000 | 9 |
| 6000-7000 | 7 |
| 7000-8000 | 6 |
| 8000-9000 | 3 |
| 9000-10000 | 1 |
| 10000-11000 | 1 |
Modal class = 4000-5000 ($f_1 = 18$)
$l = 4000$, $f_1 = 18$, $f_0 = 4$, $f_2 = 9$, $h = 1000$
Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:
$$= 4000 + \frac{18 - 4}{2(18) - 4 - 9} \times 1000$$$$= 4000 + \frac{14}{23} \times 1000$$
$$= 4000 + 608.7$$
$$= 4608.7 \text{ (about)}$$
∴ Mode ≈ 4608.7 runs
6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data. #
Answer
| Number of cars | $f_i$ |
|---|---|
| 0-10 | 7 |
| 10-20 | 14 |
| 20-30 | 13 |
| 30-40 | 12 |
| 40-50 | 20 |
| 50-60 | 11 |
| 60-70 | 15 |
| 70-80 | 8 |
Modal class = 40-50 ($f_1 = 20$)
$l = 40$, $f_1 = 20$, $f_0 = 12$, $f_2 = 11$, $h = 10$
Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:
$$= 40 + \frac{20 - 12}{2(20) - 12 - 11} \times 10$$$$= 40 + \frac{8}{17} \times 10$$
$$= 40 + 4.7$$
$$= 44.7 \text{ (about)}$$
∴ Mode = 44.7 cars