Question

1. IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the x- score that corresponds to a z- score of 2.33.

a. 142.35

b. 139.55

c. 125.95

d. 134.95

2. The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.95° F. What temperature represents the 95th percentile?

a. 99.82° F

b. 100.46° F

c. 100.16° F

d. 97.04° F

Answer 1

Solution :

Given that ,

1)

mean = = 100

standard deviation = = 15

Using standard normal table ,

z = 2.33

Using z-score formula,

x = z * +

x = 2.33 * 15 + 100 = 134.95

option d. is correct

2)

mean = = 98.6

standard deviation = = 0.95

Using standard normal table ,

P(Z < z) =95%

P(Z < 1.645) = 0.95

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 * 0.95 + 98.6 = 100.16

95th percentile = 100.160F

option c. is correct