The problem provides a table showing the time it takes for each of 40 students to travel to school. The table gives the time intervals and the corresponding frequencies. The first part asks to calculate an estimate of the mean travel time. The second part requests to draw a histogram.

1. Problem Description

The problem provides a table showing the time it takes for each of 40 students to travel to school. The table gives the time intervals and the corresponding frequencies.

The first part asks to calculate an estimate of the mean travel time. The second part requests to draw a histogram.

2. Solution Steps

(i) Calculate an estimate of the mean.

To calculate the estimated mean, we first need to find the midpoint of each time interval.

Midpoint of 0 <= m < 10 is $(0+10)/2 = 5$

Midpoint of 10 <= m < 25 is $(10+25)/2 = 17.5$

Midpoint of 25 <= m < 40 is $(25+40)/2 = 32.5$

Midpoint of 40 <= m < 50 is $(40+50)/2 = 45$

Then, we multiply each midpoint by its corresponding frequency.

$5 \times 3 = 15$

$17.5 \times 18 = 315$

$32.5 \times 15 = 487.5$

$45 \times 4 = 180$

Sum the products: $15 + 315 + 487.5 + 180 = 997.5$

Divide the sum by the total number of students (40) to find the estimated mean.

Estimated mean $= 997.5 / 40 = 24.9375$

(ii) Draw a histogram to show the information in the table.

We are given Frequency Density (FD) values for the histogram:

0 <= m < 10, FD = 0.3

10 <= m < 25, FD = 1.2

25 <= m < 40, FD = 1

40 <= m < 50, FD = 0.2

Frequency Density = Frequency / Class width

Therefore, the height of the bar representing 0 <= m < 10 is 0.

3. The width is 10-0 =

1

0. The height of the bar representing 10 <= m < 25 is 1.

2. The width is 25-10 =

1

5. The height of the bar representing 25 <= m < 40 is

1. The width is 40-25 =

1

5. The height of the bar representing 40 <= m < 50 is 0.

2. The width is 50-40 =

1

0. The problem is to find the estimate mean but here are all the steps followed by the student:

The student found the midpoint for each interval.

Multiply the midpoint with corresponding frequency

Added all the products together and divided by the total number of students which is

4

0. Also, Frequency Density (FD) is found correctly as follows:

For 0 <= m < 10: FD = 3/10 = 0.3

For 10 <= m < 25: FD = 18/15 = 1.2

For 25 <= m < 40: FD = 15/15 = 1.0

For 40 <= m < 50: FD = 4/10 = 0.4, however, it has been written as 0.

2.

3. Final Answer

Estimated Mean Travel Time: 24.9375 minutes