WAEC 2011 · Paper 2 · Q11

1. Find $x$ first. The total number of throws is $2 + 5 + x + 11 + 9 + 10 = 37 + x$, and $P(3) = \frac{x}{37 + x} = 0.26$.
2. So $x = 0.26(37 + x) = 9.62 + 0.26x$, which gives $0.74x = 9.62$ and $x = 13$.
3. The die was thrown 50 times.

(a)

1. With 50 scores, the median is halfway between the 25th and 26th.
2. The running totals are $2, 7, 20, 31, \ldots$: the 21st to 31st scores are all 4, so the 25th and 26th are both 4.
3. Median $= 4$.

(b)

1. $\sum fx = 2 + 10 + 39 + 44 + 45 + 60$
   $= 200$, so the mean is $\frac{200}{50} = 4$.
2. Then $\sum f(x - 4)^2 = 2(9) + 5(4) + 13(1) + 11(0) + 9(1) + 10(4)$
   $= 18 + 20 + 13 + 0 + 9 + 40$
   $= 100$.
3. Standard deviation $= \sqrt{\frac{100}{50}}$
   $= \sqrt 2$
   $\approx 1.41$.