Question

20) A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = .05. Suppose a random sample of 100 male students is selected and the GPA for each student is calculated. Find the interval that contains 95.44 percent of the sample means for male students. µ = Population Mean σ =Population Standard Deviation

Answer 1

Solution:-

Given that,

mean = = 3.5

standard deviation = = 0.05

n = 100

= = 3.5

= / n = 0.05 / 100 = 0.005

Using standard normal table,

P( -z < Z < z) = 0.9544

= P(Z < z) - P(Z <-z ) = 0.9544

= 2P(Z < z) - 1 = 0.9544

= 2P(Z < z) = 1 + 0.9544

= P(Z < z) = 1.9544 / 2

= P(Z < z) = 0.9772

= P(Z < 2.00) = 0.9772

= z  ± 2.00

Using z-score formula  

= z * +

= -2.00 * 0.005 + 3.5

= 3.49

Using z-score formula  

= z * +

= 2.00 * 0.005 + 3.5

= 3.51

95.44% interval = ( 3.49, 3.51 )