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Answer :
The proportion of people with IQ scores above 97.8 is 0.5714 or 57.14%.
The IQ score that falls in the lowest 25% of the distribution is 89.88.
The IQ score needed to be in the highest 5% of the distribution is 124.68.
In order to find the proportion of people aged 20 to 34 who have IQ scores above 97.8,
we have to use a standard normal distribution table.
Converting the IQ score to a z-score, we get,
⇒ z = (97.8 - mean) / standard deviation
Assume a mean IQ score of 100 and a standard deviation of 15 (as is typical for IQ tests), we get,
⇒ z = (97.8 - 100) / 15
= -0.18
Using a standard normal distribution table,
We can find that the proportion of people with IQ scores above 97.8 is 0.5714 or 57.14%.
To find the IQ score that falls in the lowest 25% of the distribution,
we have to find the z-score that corresponds to the 25th percentile. Using a standard normal distribution table,
we get a z-score of -0.675.
We can then convert this back to an IQ score,
⇒ IQ score = mean + (z-score x standard deviation)
⇒ IQ score = 100 + (-0.675 x 15)
= 89.875
So the IQ score that falls in the lowest 25% of the distribution is 89.88.
Now to find the IQ score needed to be in the highest 5% of the distribution,
we have to find the z-score that corresponds to the 95th percentile. Using a standard normal distribution table,
We get a z-score of 1.645. We can then convert this back to an IQ score,
⇒ IQ score = mean + (z-score x standard deviation)
⇒ IQ score = 100 + (1.645 x 15)
= 124.68
So the IQ score needed to be in the highest 5% of the distribution is 124.68.
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