Question

Forty percent of students at a small college have brought their own personal computer to campus. A random sample of 120 entering freshmen was taken.

a. What is the standard error of the sample proportion bringing their own computer to campus.

b. What is the probability that the sample proportions is less than 0.33?

c. What is the probability that the sample proportion is between 0.38 and 0.46?

Answer 1

Solution

(a)

Standard error = =  [p ( 1 - p ) / n] =   [(0.40 * 0.60) / 120 ] = 0.04472

(b)

P( < 0.33) =

= P[( - ) / < (0.33 - 0.40) / 0.04472]

= P(z < 1.57)

= 0.9418

(c)

= P[(0.38 - 0.40) / 0.04472 < ( - ) / < (0.46 - 0.40) / 0.04472]

= P(-0.45 < z < 1.34)

= P(z < 1.34) - P(z < -0.45)

= 0.5835