The table shows the distribution of the heights of a group of people. Height (m) 0.4 - 0.5 0.6 - 0.9 1.0 - 1.2 1.3 - 1.4 1.5 - 1.7 Number of people 2 8 12 6...
Question 1 Report
The table shows the distribution of the heights of a group of people.
| Height (m) |
0.4 - 0.5 | 0.6 - 0.9 | 1.0 - 1.2 | 1.3 - 1.4 | 1.5 - 1.7 |
| Number of people | 2 | 8 | 12 | 6 | 6 |
(a) Draw a histogram to illustrate the distribution.
(b) Using an assumed mean of 1.1m, find, correct to one decimal place, the mean height of the group.
Answer Details
(a) Histogram. The classes have unequal widths, so the bars must be drawn using frequency density, not raw frequency:
\[\text{frequency density}=\frac{\text{frequency}}{\text{class width}}\]First close the gaps between successive classes to obtain the true class boundaries, then compute the width and frequency density of each class.
| Height (m) | Class boundaries | Width | Frequency | Freq. density |
| 0.4 - 0.5 | 0.35 - 0.55 | 0.2 | 2 | 10 |
| 0.6 - 0.9 | 0.55 - 0.95 | 0.4 | 8 | 20 |
| 1.0 - 1.2 | 0.95 - 1.25 | 0.3 | 12 | 40 |
| 1.3 - 1.4 | 1.25 - 1.45 | 0.2 | 6 | 30 |
| 1.5 - 1.7 | 1.45 - 1.75 | 0.3 | 6 | 20 |
Plot frequency density on the vertical axis against height (class boundaries) on the horizontal axis, drawing the bars with no gaps between them:
(b) Mean height (assumed-mean method, \(A=1.1\) m). Use the class mid-values \(x\) and the deviation \(d=x-A\):
\[\bar{x}=A+\frac{\sum fd}{\sum f}\]| Height (m) | Mid-value \(x\) | Frequency \(f\) | \(d=x-1.1\) | \(fd\) |
| 0.4 - 0.5 | 0.45 | 2 | -0.65 | -1.30 |
| 0.6 - 0.9 | 0.75 | 8 | -0.35 | -2.80 |
| 1.0 - 1.2 | 1.10 | 12 | 0.00 | 0.00 |
| 1.3 - 1.4 | 1.35 | 6 | 0.25 | 1.50 |
| 1.5 - 1.7 | 1.60 | 6 | 0.50 | 3.00 |
| Total | 34 | 0.40 |
Therefore
\[\bar{x}=1.1+\frac{0.40}{34}=1.1+0.0118=1.1118\text{ m}\] \[\boxed{\bar{x}\approx 1.1\text{ m (to 1 decimal place)}}\]The mean height of the group is 1.1 m.
Download The App On Google Playstore
Everything you need to excel in your exams
