The Normal Distribution

Question

IQ is normally distributed with a mean of 100 and a standard deviation of 15.

1. a) Suppose one individual is randomly chosen. Find the probability that this person has an IQ greater than 95.
2. Write your answer in percent form. Round to the nearest tenth of a percent.
3. P�(IQ greater than 95) = %
4. b) Suppose one individual is randomly chosen. Find the probability that this person has an IQ less than 125.
5. Write your answer in percent form. Round to the nearest tenth of a percent.
6. P�(IQ less than 125) = %
7. c) In a sample of 700 people, how many people would have an IQ less than 110?
8. people

Answer 1

Hi Tra'von,

First Question:

Use the classic formula: z=(x-mu)/sigma

x=value given

mu=mean

sigma=standard deviation

For our problem:

x=95

mu=100

sigma=15

z=(95-100)/15

z= -0.33

From z-table:

P(Z< -0.33)=0.3707

But you were asked for greater, so apply the Complement Rule aka the "One Minus Trick:"

P(Z> -0.33)= 1-0.3707

P=0.6293

Second Question:

0.6293*100=62.93%; to nearest tenth:

62.9%

Third Question:

Same answer as second, 62.9%.

Fourth Question:

Again, use the statistical classic:

z=(x-mu)/sigma

x=125

mu=100

sigma=15

z=(125-100)/15

z=1.67

P(Z<1.67)=0.9525

Fifth Question:

0.9525*100=95.25; to nearest tenth:

95.3%

Sixth Question:

Same as fifth; 95.3%

Seventh/Eighth Question (appear to be asking for same value):

First, apply the classic yet again:

z=(x-mu)/sigma

x=110

mu=100

sigma=15

z=(110-100)/15

z=0.67

P(Z<0.67)=0.7486

Now, multiply that by your sample size, 700 people:

0.7486*700= 524.02 or, rounded to next whole number, 525.

I hope this helps.