The table shows the scores obtained when a fair die was thrown a number of...
MATHEMATICS
WAEC 2011
The table shows the scores obtained when a fair die was thrown a number of times.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
If the probability of obtaining a 3 is 0.26, find the (a) median
(b) standard deviation of the distribution.
Explanation
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
P(3) = 0.26
\(\frac{x}{37 + x} = 0.26\)
\(x = 0.26(37 + x) \implies x = 9.62 + 0.26x\)
\(x - 0.26x = 9.62 \implies 0.74x = 9.62\)
\(x = \frac{9.62}{0.74} = 13\)
Total toss frequency = 37 + x
= 37 + 13
= 50.
(a) Median = \(\frac{n + 1}{2}\)
= \(\frac{50 + 1}{2}\)
= 25.5
From the table, 25.5 occurs at 4 which is the median.
(b) Standard deviation = \(\sqrt{\frac{\sum fd^{2}}{\sum f}}\)
| \(x\) | \(f\) | \(\fx\) | \(d = |x - \bar{x}|\) | \(d^{2}\) | \(fd^{2}\) |
| 1 | 2 | 2 | -3 | 9 | 18 |
| 2 | 5 | 10 | -2 | 4 | 20 |
| 3 | 13 | 39 | -1 | 1 | 13 |
| 4 | 11 | 44 | 0 | 0 | 0 |
| 5 | 9 | 45 | 1 | 1 | 9 |
| 6 | 10 | 60 | 2 | 4 | 40 |
| \(\sum\) | 50 | 200 | 100 |
\(Mean(\bar{x}) = \frac{\sum fx}{\sum f}\)
= \(\frac{200}{50}
= 4
S.D = \(\sqrt{\frac{100}{50}}\)
= \(\sqrt{2}\)
= 1.4142
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