Question

80% of students at Merrimack College live on campus (normally— not during a pandemic). 120 randomly selected students at Merrimack are surveyed and asked their living situation. Let p̂ be the proportion of students in the sample survey who report they live on campus.

a. Find the mean and standard deviation of the sampling distribution of p̂.

b. Use normalcdf with the sample proportion p̂ to determine the probability that at least 75% of the students in the survey live on campus.

Answer 1

Solution

Given that,

p = 0.80

1 - p = 1 - 0.80 = 0.20

n = 120

a)   = p = 0.80

=  [p( 1 - p ) / n] = [(0.80 * 0.20) / 120 ] = 0.0365

b) P(    0.75 ) = 1 - P( 0.75 )

= 1 - P(( - ) /    (0.75 - 0.80) / 0.0365)

= 1 - P(z -1.37)

Using z table

= 1 - 0.0853

= 0.9147