Question 3.29: Determine the mean value for the heights of the 200 adults u...
Determine the mean value for the heights of the 200 adults using the data in the table in Example 3.26.
The values for each individual class are best found by producing a table using the class midpoints and frequencies, remembering that the class midpoint is found by dividing the sum of the upper and lower class boundaries by 2. For example, the mean value for the first class interval is \frac{149.5+154.5}{2}=152. The completed table is shown below.
| Midpoint (x) of height (cm) |
Frequency (f) | fx |
| 152 | 4 | 608 |
| 175 | 9 | 1413 |
| 162 | 15 | 2430 |
| 167 | 21 | 3507 |
| 172 | 32 | 5504 |
| 177 | 45 | 7965 |
| 182 | 41 | 7462 |
| 187 | 22 | 4114 |
| 192 | 9 | 1728 |
| 197 | 2 | 394 |
| Total | ∑f = 200 | ∑fx = 35,125 |
We hope you can see how each of the values was obtained. When dealing with relatively large numbers, be careful with your arithmetic, especially when you are keying variables into your calculator!
Now that we have the required total, the mean value of the distribution can be found:
\begin{aligned} \text{Mean value} \quad \bar{x} &=\frac{\sum f x}{\sum f}=\frac{35,125}{200} \\&=175.625 \pm 0.5 \ cm\end{aligned}
Notice that our mean value of heights has the same margin of error as the original measurements. The value of the mean cannot be any more accurate than the measured data from which it was obtained!