(a) The distribution of the lives (in days) of 40 transitor batteries is shown in
FURTHER MATHEMATICS
WAEC 2009
(a) The distribution of the lives (in days) of 40 transitor batteries is shown in the table:
Battery life (in days) | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |
| Frequency | 4 | 7 | 13 | 8 | 6 | 2 |
(a) Draw a histogram for the distribution.
(b) Use your graph in (a) to determine the mode for the distribution.
(c) Using an assumed mean of 43 days, calculate the mean of the distribution.
(d) What percentage number of batteries will live for less than 31 days or more than 45 days?
Explanation
| Battery life (in days) | Freq | Class boundaries |
| 26 - 30 | 4 | 25.5 - 30.5 |
| 31 - 35 | 7 | 30.5 - 35.5 |
| 36 - 40 | 13 | 35.5 - 40.5 |
| 41 - 45 | 8 | 40.5 - 45.5 |
| 46 - 50 | 6 | 45.5 - 50.5 |
| 51 - 55 | 2 | 50.5 - 55.5 |
| 40 |
.jpg)
(b) Mode = 38.4
(c) Assumed mean, A = 43 days
Battery life (in days) | Mid-mark (x) | Frequency (f) | \(d = x - A\) | \(fd\) |
| 26 - 30 | 28 | 4 | -15 | -60 |
| 31 - 35 | 33 | 7 | -10 | -70 |
| 36 - 40 | 38 | 13 | -5 | -65 |
| 41 - 45 | 43 | 8 | 0 | 0 |
| 46 - 50 | 48 | 6 | 5 | 30 |
| 51 - 55 | 53 | 2 | 10 | 20 |
| 40 | -145 |
Mean \(\bar{x} = A + \frac{\sum fd}{\sum f}\)
= \(43 + \frac{-145}{40}\)
= \(43 - 3.625\)
= \(39.375\)
(d) Number of batteries that will live for less than 31 days or more than 45 days = 4 + 6 + 2 = 12
\(%age = \frac{12}{40} \times 100% = 30%\)
Post an Explanation Or Report an Error
If you see any wrong question or answer, please leave a comment below and we'll take a look. If you doubt why the selected answer is correct or need additional more details? Please drop a comment or Contact us directly. Your email address will not be published. Required fields are marked *