Question

A sample of final exam scores is normally distributed with a mean equal to 24 and a variance equal to 25. Part (a) What percentage of scores are between 19 and 29? (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 18? (Round your answer to four decimal places.) Part (d) What is the probability of a score less than 30? (Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 24

variance = ² = 25

standard deviation = = ² = 25 = 5

a) P(19 < x < 29) = P[(19 - 24)/ 5 ) < (x - ) /  < (29 - 24) / 5) ]

= P(-1.00 < z < 1.00)

= P(z < 1.00) - P(z < -1.00)

Using z table,

= 0.8413 - 0.1587

= 0.6826

The percentage is = 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 + 24

x = 30.4

c) P(x < 18) = P[(x - ) / < (18 - 24) / 5]

= P(z < -1.20)

Using z table,

= 0.1151

d) P(x < 30) = P[(x - ) / < (30 - 24) / 5]

= P(z < 1.20)

Using z table,

= 0.8849