Question

The time taken to complete an assignment in a class follows a normal distribution with a mean of105 minutes and a standard deviation of 10 minutes.

A. What is the probability that a randomly selected student takes between 100and 112 minutes to complete the exam?

B. What is the probability that a randomly selected student takes less than 115minutes to complete the exam?

C. What is the time that corresponds to the 38th percentile?

D. Between which two values does the middle 74.25% of the area underthis normal curve lie?

Answer 1

Solution :

Given that ,

mean = = 105

standard deviation = = 10

A) P(100 < x < 112 ) = P[(100 - 105)/ 10) < (x - ) /  < (112 - 105) / 10) ]

= P(-0.50 < z < 0.70)

= P(z < 0.70) - P(z < -0.50)

Using z table,

= 0.7580 - 0.3085

= 0.4495

B) P(x < 115)

= P[(x - ) / < (115 - 105) / 10]

= P(z < 1.00)

Using z table,

= 0.8413

C) Using standard normal table,

P(Z < z) = 38%

= P(Z < z ) = 0.38

= P(Z < -0.31 ) = 0.38  

z = -0.31

Using z-score formula,

x = z * +

x = -0.31 * 10 + 105

x = 101.9 minutes

D) Using standard normal table,

P( -z < Z < z) = 74.25%

= P(Z < z) - P(Z <-z ) = 0.7425

= 2P(Z < z) - 1 = 0.7425

= 2P(Z < z) = 1 + 0.7425

= P(Z < z) = 1.7425 / 2

= P(Z < z) = 0.87125

= P(Z < 1.13) = 0.87125

= z  ± 1.13

Using z-score formula,

x = z * +

x = -1.13 * 10 + 105

x = 93.7 minutes

Using z-score formula,

x = z * +

x = 1.13 * 10 + 105

x = 116.3 minutes

The middle 74.25% are from 93.7 minutes to 116.3 minutes