NCERT Solutions: Statistics (Exercise 13.2)

Table of Contents
1. Q1: 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.
2. Q2: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: Determine the modal lifetimes of the components.
3. Q3: 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:
4. Q4: 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.
5. Q5: 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.
6. Q6: 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:
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Q1: 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.

Sol: Mode:

Here, the highest frequency is 23.

The frequency 23 corresponds to the class interval 35 - 45.

∴ The modal class is 35 - 45.

Class size (h) = 10.

Lower limit (l) = 35.

Frequency of the modal class (f₁) = 23.

Frequency of the class preceding the modal class (f₀) = 21.

Frequency of the class succeeding the modal class (f₂) = 14.

Use the formula Mode = l + (f₁ - f₀) / (2f₁ - f₀ - f₂) × h.

Substitute values: Mode = 35 + (23 - 21) / (2·23 - 21 - 14) × 10.

Therefore, Mode = 35 + 2/11 × 10 = 35 + 20/11 = 36.818...

Mode ≈ 36.82 years.

Mean

Let assumed mean a = 40.

Class width h = 10.

Apply the assumed mean method (compute d_i = (x_i - a)/h, form f_id_i and sum).

Carrying out the table calculations (as in the provided working) gives the required value.

Mean = 35.37 years.

Comparison and interpretation

The mean (35.37 years) represents the average age of admitted patients, while the mode (≈ 36.82 years) represents the most frequently occurring age (the modal value) in the grouped data.

The two measures are close in value, indicating the distribution is fairly near symmetric with only a slight skew; since the mean is a little less than the mode, this suggests a slight negative (left) skew in the age distribution.

Q2: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Sol: Here, the highest frequency = 61.

∴ The modal class = 60 - 80.

l = 60.

h = 20.

f₁ = 61.

f₀ = 52.

f₂ = 38.

Use the formula Mode = l + (f₁ - f₀) / (2f₁ - f₀ - f₂) × h.

Substitute values: Mode = 60 + (61 - 52) / (2·61 - 52 - 38) × 20.

Therefore, Mode = 60 + 9/32 × 20 = 60 + 180/32 = 60 + 5.625.

Modal lifetime = 65.625 hours.

Q3: 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:

Sol: Mode: 

∵ The maximum number of families 40 have their total monthly expenditure in interval 1500 - 2000.

∴ Modal class is 1500 - 2000.

l = 1500.

h = 500.

f₁ = 40.

f₀ = 24.

f₂ = 33.

Use the formula Mode = l + (f₁ - f₀) / (2f₁ - f₀ - f₂) × h.

Substitute values: Mode = 1500 + (40 - 24) / (2·40 - 24 - 33) × 500.

Therefore, Mode = 1500 + 16/23 × 500 = 1500 + 8000/23 = 1847.826...

Modal monthly expenditure ≈ Rs 1847.83.

Mean: 

Let assumed mean a = 3250.

Class width h = 500.

Form the table of class mid-points x_i, compute d_i = (x_i - a)/h, then f_id_i, and sum f_id_i as shown in the working.

Carrying out the computations yields the required mean.

Mean monthly expenditure = Rs 2662.50.

Q4: 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.

Sol: Mode: 

Since the class 30 - 35 has the greatest frequency and h = 5.

l = 30.

f₁ = 10.

f₀ = 9.

f₂ = 3.

Use the formula Mode = l + (f₁ - f₀) / (2f₁ - f₀ - f₂) × h.

Substitute values: Mode = 30 + (10 - 9) / (2·10 - 9 - 3) × 5.

Therefore, Mode = 30 + 1/8 × 5 = 30 + 0.625.

Mode ≈ 30.625.

Mean:

Let the assumed mean a = 37.5.

Since, h = 5.

Form the table with class mid-points, compute d_i = (x_i - a)/h, multiply by frequencies, sum f_id_i, and apply assumed mean formula to obtain the mean.

Thus, the required mean is 29.2.

Interpretation

The mean (29.2) gives the average teacher-student ratio across the states represented, while the mode (≈ 30.63) gives the most commonly observed ratio class.

The mean being slightly less than the mode suggests a small negative (left) skew in the distribution of ratios; the two measures are reasonably close, so no extreme skewness is indicated.

Q5: 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.

Sol: The class 4000 - 5000 has the highest frequency i.e., 18.

∴ h = 1000.

l = 4000.

f₁ = 18.

f₀ = 4.

f₂ = 9.

Use the formula Mode = l + (f₁ - f₀) / (2f₁ - f₀ - f₂) × h.

Substitute values: Mode = 4000 + (18 - 4) / (2·18 - 4 - 9) × 1000.

Therefore, Mode = 4000 + 14/23 × 1000 = 4000 + 608.695...

Mode ≈ 4608.7.

Q6: 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:

Sol: ∵ The class 40 - 50 has the maximum frequency i.e., 20.

∴ f₁ = 20.

f₀ = 12.

f₂ = 11.

h = 10.

l = 40.

Use the formula Mode = l + (f₁ - f₀) / (2f₁ - f₀ - f₂) × h.

Substitute values: Mode = 40 + (20 - 12) / (2·20 - 12 - 11) × 10.

Therefore, Mode = 40 + 8/17 × 10 = 40 + 80/17 = 44.705...

Mode ≈ 44.7.