Question

scores on a college entrance test are normally distributed with a mean 300 and a standard deviation of 50
if a test score is picked at random what is the probability that the score is less than 215 or more than 345

b) find two test scores that divide the normal curve into a middle of 0.92 and two 0.04 areas

Answer 1

Solution :

mean = = 300

standard deviation = = 50

Using standard normal table,

(a)

P(x < 215) = P((x - ) / < (215- 300) / 50) = P(z < -1.7)

Using standard normal table,

P(x < 215) = 0.0446

P(x > 345) = 1 - P(x < 345)

= 1 - P((x - ) / < (345 - 300) / 50)

= 1 - P(z < 0.9)

= 1 - 0.8159   

= 0.1841

P(x > 345) = 0.1841

The probability that the score is less than 215 or more than 345 is = 0.0446 + 0.1841 = 0.2301

(b)

P(Z > z) = 0.92

1 - P(Z < z) = 0.92

P(Z < z) = 1 - 0.92 = 0.08

P(Z < -1.405) = 0.08

z = -1.405

Using z-score formula,

x = z * +

x = -1.405 * 50 + 300 = 229.75

P(Z < z) = 0.92

P(Z < 1.405) = 0.92

z = 1.405

Using z-score formula,

x = z * +

x = 1.405 * 50 + 300 = 370.25

Two test scaore are : 229.75 and 370.25