Question

A college research group reported that 43​% of college students aged​ 18-24 would spend their spring breaks relaxing at home in 2009. A sample of 165 college students was selected. Complete parts a through d below.

a. Calculate the standard error of the proportion.

σp = ​(Round to four decimal places as​ needed.)

b. What is the probability that less than​ 40% of the college students from the sample spent their spring breaks relaxing at​ home? ​

P(Less than​ 40% of the college students from the sample spent their spring breaks relaxing at ​home) = (Round to four decimal places as​ needed.)

c. What is the probability that more than​ 50% of the college students from the sample spent their spring breaks relaxing at​ home?

​P(More than​ 50% of the college students from the sample spent their spring breaks relaxing at ​home) = ​(Round to four decimal places as​ needed.)

d. What is the probability that between 44​% and 54​% of the college students from the sample spent their spring breaks relaxing at​ home? ​

P(Between 44​% and 54​% of the college students from the sample spent their spring breaks relaxing at ​home) = ​(Round to four decimal places as​ needed.)

Answer 1

Solution

Given that,

p = 0.43

1 - p = 1 - 0.43 = 0.57

n = 165

a)   = p = 0.43

=  [p ( 1 - p ) / n] =   [(0.43 * 0.57) / 165 ] = 0.0385

b) P( < 0.40)

= P[( - ) / < (0.40 - 0.43) / 0.0385]

= P(z < -0.78)

Using z table,

= 0.2177

c) P( > 0.50) = 1 - P( < 0.50)

= 1 - P(( - ) / < ( 0.50 - 0.43) / 0.0385)

= 1 - P(z < 1.82)

Using z table

= 1 - 0.9656

= 0.0344

d) P( 0.44 < < 0.54 )

= P[(0.44 - 0.43) / 0.0385 < ( - ) / < (0.54 - 0.43) / 0.0385 ]

= P(0.26 < z < 2.86)

= P(z < 2.86) - P(z < 0.26)

Using z table,   

= 0.9979 - 0.6026

= 0.3953