The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows : 50 99 81 86 69 85 93 63 92 65 77 74 76 71 ...
Question 1 Report
The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows :
50 99 81 86 69 85 93 63 92 65 77 74 76 71 90 74 81 94 67 75 95 81 68 105 99 68 75 75 76 73 79 74 80 69 74 62 74 80 79 68 79 75 75 71 83 75 80 85 81 82
(a) Construct a frequency distribution table, using class intervals 45 - 54, 55 - 64, etc.
(b) Draw the histogram for the distribution
(c) Use your histogram to estimate the mode.
(d) Calculate the mean number of births.
Answer Details
(a) Frequency distribution
| Number of births | Class midpoint, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|
| 45–54 | 49.5 | 1 | 49.5 |
| 55–64 | 59.5 | 2 | 119.0 |
| 65–74 | 69.5 | 15 | 1042.5 |
| 75–84 | 79.5 | 21 | 1669.5 |
| 85–94 | 89.5 | 7 | 626.5 |
| 95–104 | 99.5 | 3 | 298.5 |
| 105–114 | 109.5 | 1 | 109.5 |
| Total | 50 | 3915 |
(b) Histogram
The continuous class boundaries are \(44.5\text{–}54.5,\ 54.5\text{–}64.5,\ldots,\ 104.5\text{–}114.5\). Since all class widths are 10, the heights of the adjacent bars are their frequencies.
(c) Estimated mode
The modal class is \(75\text{–}84\). From the histogram, or by interpolation within this class,
\[\text{Mode}=L+\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\times c\]
\[=74.5+\frac{21-15}{(21-15)+(21-7)}\times10\]
\[=74.5+\frac{6}{20}\times10=\boxed{77.5\text{ births (approximately)}}.\]
(d) Mean number of births
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3915}{50}=\boxed{78.3\text{ births}}.\]
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