Question

A sample of test scores is normally distributed with a mean equal to 22 and a variance equal to 25.

Part (a) What percentage of scores are between 17 and 27? (Round your answer to two decimal places.)

____%

Part (b) What raw score is the cutoff for the top 10% of scores? (Round your answer to one decimal place.)

Part (c) What is the proportion below 15? (Round your answer to four decimal places.)

Part (d) What is the probability of a score less than 29? (Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 22

standard deviation = = ² = 25 = 5

a) P(17 < x < 27) = P[(17 - 22)/ 5 ) < (x - ) /  < (27 - 22) / 5) ]

= P(-1.0 < z < 1.0 )

= P(z < 1.0) - P(z < -1.0 )

Using z table,

= 0.8413 - 0.1587

= 0.6826

percentage = 68.26%

b) Using standard normal table,

P(Z > z) = 10%

= 1 - P(Z < z) = 0.10

= P(Z < z) = 1 - 0.10

= P(Z < z ) = 0.90

= P(Z < 1.28 ) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 5 + 22

x = 28.4

c) P(x < 15)

= P[(x - ) / < (15 - 22) / 5]

= P(z < -1.40 )

Using z table,

= 0.0808

d) P(x < 29)

= P[(x - ) / < (29 - 22) / 5]

= P(z < 1.40 )

Using z table,

= 0.9192