# A normal distribution of IQ scores

1. Oct 19, 2015

### toothpaste666

1. The problem statement, all variables and given/known data
It is known that the IQ score of ten-year-old children in a particular population has a normal distribution with mean 100 and standard deviation 15.

(a) What proportion of this population have an IQ score above 115?

(b) Mary’s IQ is equal to the 80th percentile of this population. What does her IQ score equal?

3. The attempt at a solution
part a)
z = (115 - 100)/15 = 15/15 = 1
we want
1 - F(1)
using the tables for the standard normal distribution F(1) = .8413
1 - .8413 = .1587

b) F(z) = .8
it seems that the closest to .8 I can find in the table is .7995 which corresponds to z = .84
z = .84 = (X - 100)/15
X - 100 = 12.6
X = 112.6

i am not sure if this is wrong because it is not the exact 80th percentile and I am not sure if I should round up to 113 or something or just leave it the way it is. Any feedback would be greatly appreciated

2. Oct 19, 2015

### SteamKing

Staff Emeritus
You can always interpolate the standard normal table for F(z) = 0.8 to find z, if this value, 0.8, does not appear directly tabulated.

3. Oct 19, 2015

### toothpaste666

how does this work?

4. Oct 19, 2015

### SteamKing

Staff Emeritus
5. Oct 19, 2015

### Staff: Mentor

The above looks OK, although I didn't check your numbers. I would write it a little differently though.
Here the random variable X = the IQ score
$Pr(X > 115) = Pr( (X - 100)/15 > (115 - 100)/15) = Pr(Z > 1) \approx 1 - .8413 = .1587$
Take the two probability values that surround .8 and interpolate the two associated z values.

6. Oct 20, 2015

### toothpaste666

so .7995 corresponds to z = .84 and the next one .8023 corresponds to z=.85 by linear interpolation
z = [(.85-.84)/(.8023-.7995)](.8-.7995) + .84 = .842 (approximately)

.842 = (X-100)/15
12.63 = X - 100
X = 112.63

7. Oct 20, 2015

### Staff: Mentor

Which you can round to 113, since it's an IQ score.