Question

IQ is normally distributed with a mean of 100 and a standard deviation of 15. a) Suppose one individual is randomly chosen. Find the probability that this person has an IQ greater than 95. Write your answer in percent form. Round to the nearest tenth of a percent. P (IQ greater than 95) = 63.1 Correct% b) Suppose one individual is randomly chosen. Find the probability that this person has an IQ less than 125. Write your answer in percent form. Round to the nearest tenth of a percent. P (IQ less than 125) = 95.2 Correct% c) In a sample of 700 people, how many people would have an IQ less than 110? 523.25 Correctpeople d) In a sample of 700 people, how many people would have an IQ greater than 140? 697.35 Incorrectpeople

Answer 1

Answer:

(a) P(X > 95) = 63.1%

(b) P(X < 125) = 95.2%

(c) 523.25 people would have an IQ less than 110.

(d) 697.2 people would have an IQ greater than 140.

Step-by-step explanation:

We are given that IQ is normally distributed with a mean of 100 and a standard deviation of 15.

Let X = IQ level of a person

So, X ~ N($\mu=100,\sigma^{2} = 15^{2}$)

The z score probability distribution is given by;

                Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)  

where, $\mu$ = population mean

           $\sigma$ = population standard deviation

(a) Probability that this person has an IQ greater than 95 is given by = P(X > 95)

P(X > 95) = P( $\frac{X-\mu}{\sigma}$ > $\frac{95 - 100}{15}$ ) = P(Z > -0.33) = P(Z < 0.33) = 0.631 or 63.1%

(b) Probability that this person has an IQ less than 125 is given by = P(X < 125)

   P(X < 125) = P( $\frac{X-\mu}{\sigma}$ < $\frac{125 - 100}{15}$ ) = P(Z < 1.67) = 0.9525 or 95.2%

(c) Now as we are given with the sample of 700 people. So, the z score probability distribution is now given as;

                    Z = $\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)  

where, $\mu$ = population mean

           $\sigma$ = population standard deviation

           n = sample size = 700

So, Probability of people that have an IQ less than 110 = P(X < 110)

   P(X < 110) = P( $\frac{X-\mu}{\sigma}$ < $\frac{110 - 100}{15}$ ) = P(Z < 0.67) = 0.7475 or 74.75%

Now, people having an IQ less than 110 in a sample of 700 = 74.75% of 700

                 = $\frac{74.75}{100} \times 700$ = 523.25

Therefore, 523.25 people would have an IQ less than 110.

(d) Probability of people that have an IQ greater than 140 = P(X > 140)

   P(X > 140) = P( $\frac{X-\mu}{\sigma}$ > $\frac{140 - 100}{15}$ ) = P(Z > 2.67) = 0.9962 or 99.6%

Now, people having an IQ greater than 140 in a sample of 700 = 99.6% of 700

                 = $\frac{99.6}{100} \times 700$ = 697.2

Therefore, 697.2 people would have an IQ greater than 140.