Question

IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individual's IQ score is found to be 123.

A.What percentage of individuals will score above 123?

B.What percentage of individuals will score below 123?

c. What percentage of individuals will score between 123 and 100?

d. This individual was trying to be in the 80th percentile; did they achieve this? how can you tell?

e. what can we say about someone with a z score of -1.61? How many will score above? below? what is their IQ score?

f. what IQ score(s) will encompass the middle 39%?

Answer 1

Mean = = 100

Standard deviation = = 15

A)

We have to find P(X > 123)

For finding this probability we have to find z score.

That is we have to find P(Z > 1.53)

P(Z > 1.53) = 1 - P(Z < 1.53) = 1 - 0.9374 = 0.0626 ( Using z table)

Percentage = 6.26%

B)

We have to find P(X < 123)

P(X < 123) = 1 - P(Z > 123) = 1 - 0.0626 = 0.9374

Percentage = 93.74%

C)

We have to find P( 123 < X < 100) = P(100 < X < 123)

For finding this probability we have to find z score.

That is we have to find P( 0 < Z < 1.53)

P( 0 < Z < 1.53) = P(Z < 1.53) - P(Z < 0) = 0.9374 - 0.5 = 0.4374

Percentage = 43.74%

d)

We have given P(X < x) = 0.80

z value 0.80 is 0.84

We have to find value of x