Question

Q6. (15 points) IQ examination scores of 500 members of a club are normally distributed with mean of 165 and SD of 15.

Answer 1

Answer:

a) 0.8413

b) 421

c) $P_{95} = 189.675$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 165

Standard Deviation, σ = 15

We are given that the distribution of  IQ examination scores is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(IQ scores at most 180)

P(x < 180)

$P( x < 180) = P( z < \displaystyle\frac{180 - 165}{15}) = P(z < 1)$

Calculation the value from standard normal z table, we have,  

$P(x < 180) = 0.8413 = 84.13\%$

b) Number of the members of the club have IQ scores at most 180

n = 500

$\text{Members} = n\times \text{P(IQ scores at most 180)}\\= 500\times 0.8413\\=420.65 \approc 421$

c) P(X< x) = 0.95

We have to find the value of x such that the probability is 0.95

$P( X < x) = P( z < \displaystyle\frac{x - 165}{15})=0.95$  

Calculation the value from standard normal z table, we have,  

$P(z < 1.645) = 0.95$

$\displaystyle\frac{x - 165}{15} = 1.645\\\\x = 189.675$  

$P_{95} = 189.675$