Question

2. The New York Times reported that the average time to download the homepage from the IRS website was 0.8 seconds. Suppose the download time was normally distributed with standard deviation of 0.2 seconds. If random samples of 30 download times are selected,

a. what proportion of sample means will be less than 0.75 seconds

b. what proportion of the sample means will be between 0.7 and 0.9 seconds c. 90% of sample means will be less than what value?

Answer 1

Solution :

Given that ,

= 0.8

= / n = 0.2 / 30 = 0.0365

a) P( < 0.75 ) = P(( - ) / < (0.75 - 0.8) / 0.0365)

= P(z < -1.37 )

Using z table

= 0.0853

b) P(0.7 < < 0.9)  

= P[(0.7 - 0.8) / 0.0365 < ( - ) / < (0.9 - 0.8) / 0.0365)]

= P( -2.74 < Z < 2.74)

= P(Z < 2.74) - P(Z < -2.74)

Using z table,  

= 0.9969 - 0.0031

= 0.9938

c) Using standard normal table,

P(Z < z) = 90%

= P(Z < z ) = 0.90

= P(Z < 1.28 ) = 0.90

z = 1.28

Using z-score formula  

= z * +   

= 1.28 * 0.0365 + 0.8

= 0.85