Help me please with this math question...?

Question

Intelligence Scale are known to be normally distributed with a mean of 100 and a standard deviation of 16.

What score is exceeded by the highest 10% of all possible IQ scores??

Answer 1

Let X denote the random IQ of a person.

Then, P(X<=y) = P((X-100)/16<(y-100)/16) = Phi((y-100)/16), where Phi(.) denotes the cdf of the standard normal distribution.

Therefore, P(X>y), the probability of exceeding the IQ-score y, is P(X>y) = 1-Phi((y-100)/16). We would like to find the y such that this probability is 0.1.

This gives:
1-Phi((y-100)/16) = 0.1, which is the same as
Phi((y-100)/16) = 0.9.

Now, denoting the inverse cdf of the standard normal distribution by G and applying it to both sides of the equation, we get:

(y-100)/16 = G(0.9), which can be rearranged into
y = 16*G(0.9) + 100.

The inverse cdf of the std normal distribution is called NORMSINV(.) in Excel. Calling =NORMSINV(0.9) in an empty cell gives 1.2816 after rounding, so the answer is:

y = 16*1.2816+100 = 120.5156 ~ 120.5.

Answer 2

You would convert this to a standard normal distribution

Z = (X - mean)/(std dev), the z value for the 90th percentile is1.28

1.28 = (X - 100)/16, solve for X

X = 120.5 rounded

P(Z>1.28) = P(X>120.5) = 0.10

please check for errors

Answer 3

i have no idea it gives a headache just to think that you have to do it