The Intelligence Quotient (IQ) scores for adults in City A are normally distributed with a mean of 108 and a standard deviation of 10 - HSC - SSCE Mathematics Standard - Question 35 - 2020 - Paper 1

From part (a), we found that Yin's IQ corresponds to a z-score of 1 standard deviation above the mean for City B.

In City B, the mean is 112 and the standard deviation is 16, so:

$\mu = 112, \sigma = 16$

Thus, the percentage of the population in City B with an IQ score lower than Yin’s is:

• At a z-score of 1, approximately 84% of the population have an IQ score lower than Yin's.

Given that City B has 1,000,000 adults, the expected number of adults in City B with an IQ score lower than Yin’s is:

$1,000,000 \times 0.84 = 840,000$

Let Simon's IQ score be $S$. Since Simon's z-score in City A is equal to the z-score in City B, we express this mathematically.

For City A: $z = \frac{S - 108}{10}$

For City B: $z = \frac{S - 112}{16}$

Setting the two equations equal to each other:

$\frac{S - 108}{10} = \frac{S - 112}{16}$

Cross-multiply to eliminate fractions:

$16(S - 108) = 10(S - 112)$

Expanding both sides: $16S - 1728 = 10S - 1120$

Rearranging gives: $6S = 608$

Therefore, $S = \frac{608}{6} = 101.3$

So, Simon’s IQ score is approximately 101.3.