# IQ test scores are normally di

IQ test scores are normally distributed with a mean of100 and a standard deviation of 15. An individual’s IQ score isfound to be 123.

A.What percentage of individuals will score above123?

B.What percentage of individuals will score below123?

c. What percentage of individuals will score between123 and 100?

d. This individual was trying to be in the 80thpercentile; 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 IQscore?

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

Answer:

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