The problem provides a table of heights of 120 boys in an athletics club, grouped into classes. (a) We need to determine the modal class. (b) We need to estimate the mean height. (c) We need to find the probability that a randomly chosen boy has a height greater than 1.8m. (d) We need to calculate the probability that, when three boys are chosen at random, one has a height greater than 1.8m and the other two each have a height of 1.4m or less.

1. Problem Description

The problem provides a table of heights of 120 boys in an athletics club, grouped into classes.

(a) We need to determine the modal class.

(b) We need to estimate the mean height.

(c) We need to find the probability that a randomly chosen boy has a height greater than 1.8m.

(d) We need to calculate the probability that, when three boys are chosen at random, one has a height greater than 1.8m and the other two each have a height of 1.4m or less.

2. Solution Steps

(a) (i) The modal class is the class with the highest frequency. Looking at the table, the class $1.5 < h \le 1.6$ has the highest frequency of

3

0.

(a) (ii) To estimate the mean height, we first find the midpoint of each class interval.

- $1.3 < h \le 1.4$: Midpoint = $\frac{1.3 + 1.4}{2} = 1.35$

- $1.4 < h \le 1.5$: Midpoint = $\frac{1.4 + 1.5}{2} = 1.45$

- $1.5 < h \le 1.6$: Midpoint = $\frac{1.5 + 1.6}{2} = 1.55$

- $1.6 < h \le 1.7$: Midpoint = $\frac{1.6 + 1.7}{2} = 1.65$

- $1.7 < h \le 1.8$: Midpoint = $\frac{1.7 + 1.8}{2} = 1.75$

- $1.8 < h \le 1.9$: Midpoint = $\frac{1.8 + 1.9}{2} = 1.85$

Now, we multiply each midpoint by its frequency, sum these values, and divide by the total frequency (120).

Estimated mean height = $\frac{(1.35 \times 7) + (1.45 \times 18) + (1.55 \times 30) + (1.65 \times 24) + (1.75 \times 27) + (1.85 \times 14)}{120}$

Estimated mean height = $\frac{9.45 + 26.1 + 46.5 + 39.6 + 47.25 + 25.9}{120}$

Estimated mean height = $\frac{194.8}{120} = 1.62333...$

(b) (i) The number of boys with a height greater than 1.8m is

1

4. The total number of boys is

1

2

0. The probability that a randomly chosen boy has a height greater than 1.8m is $\frac{14}{120} = \frac{7}{60}$.

(b) (ii) The number of boys with height greater than 1.8m is

1

4. The probability of choosing a boy with height greater than 1.8m is $P(>1.8) = \frac{14}{120} = \frac{7}{60}$.

The number of boys with height 1.4m or less is $7+18 = 25$.

The probability of choosing a boy with height 1.4m or less is $P(\le 1.4) = \frac{25}{120} = \frac{5}{24}$.

We are choosing 3 boys. We want one boy with height greater than 1.8m and two boys with height 1.4m or less. The order matters, so we have to consider the different possible orderings. The number of different possible orderings is

3. The probability of each of the possible orderings is

$P(>1.8) \times P(\le 1.4) \times P(\le 1.4) = \frac{7}{60} \times \frac{5}{24} \times \frac{5}{24} = \frac{175}{34560}$.

The probability of one boy greater than 1.8m and two boys 1.4m or less is

$3 \times \frac{175}{34560} = \frac{525}{34560} = \frac{35}{2304}$

3. Final Answer

(a) (i) $1.5 < h \le 1.6$

(a) (ii) $1.62$

(b) (i) $\frac{7}{60}$

(b) (ii) $\frac{35}{2304}$