A normal distribution of IQ scores

[toothpaste666]

Homework Statement

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?

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

 

[SteamKing]

Homework Statement

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?

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

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.

 

[toothpaste666]

how does this work?

 

[SteamKing]

how does this work?

https://en.wikipedia.org/wiki/Linear_interpolation

This is a good tool to know when working with tabulated data.

 

[Mark44]

Homework Statement

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?

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

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
$Pr(X > 115) = Pr( (X - 100)/15 > (115 - 100)/15) = Pr(Z > 1) \approx 1 - .8413 = .1587$

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

Take the two probability values that surround .8 and interpolate the two associated z values.

 

[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

 

[Mark44]

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

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