Question

Descriptive analysis revealed that 68% of students passed Test 1. Suppose a random sample of 18 students is selected and defined ˆp as the proportion of students in the sample who passed Test 1.

Determine the mean of the sampling distribution of ˆp.
μˆp=

Determine the standard deviation of the sampling distribution of ˆp. Round the solution to six decimal places.
σˆp=

Determine the probability that between 63.3% and 77.3% of a random sample of 18 students passed Test 3. Round the solution to four decimal places.
P(0.633<ˆp<0.773) =

Determine the probability that less than 92.3% of a random sample of 18 students passed Test 3. Round the solution to four decimal places.
P(ˆp<0.923) =

Answer 1

Solution

Given that,

p = 0.68

1 - p = 1 - 0.68 = 0.32

n = 18

= p = 0.68

  [p ( 1 - p ) / n] = [(0.68 * 0.32) / 18 ] = 0.109949

a) P( 0.633 < < 0.773 )

= P[(0.633 - 0.68) / 0.109949 < ( - ) / < (0.773 - 0.68) / 0.109949]

= P(-0.43 < z < 0.85)

= P(z < 0.85) - P(z < -0.43)

Using z table,   

= 0.8023 - 0.3336

= 0.4687

b) P( < 0.923 )

= P[( - ) / < (0.923 - 0.68) / 0.109949]

= P(z < 2.21)

Using z table,

= 0.9864