Question

the dean of student life at Gonzaga University reported recently that 0.75 of all Gonzaga students...

the dean of student life at Gonzaga University reported recently that 0.75 of all Gonzaga students live on campus. Gus Spencer, a psychology professor at Gonzaga, has selected a random sample of 75 students for a behavior study.a. Calculate the standard error of the proportion. b. What is the probability the proportion of students in Gus’s study who live on campus is no greater than0.72?

Homework Answers

Answer #1

Solution

Given that,

p = 0.75

1 - p = 1 - 0.75 = 0.25

n = 75

= p = 0.75

a) =  [p ( 1 - p ) / n] =   [(0.75 * 0.25) / 75 ] = 0.05

b) P( < 0.72)

= P[( - ) / < (0.72 - 0.75) / 0.05]

= P(z < -0.60)

Using z table,

= 0.2743

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