Question

A national survey indicated that 70% of 18- to 29-year-olds would prefer to start their own business rather than work for someone else. A random sample of 600 18- to 29years-olds is obtained today.

a) (2 points) Find the mean of this sampling distribution of proportion.

b) (2 points) Find the standard deviation of this sampling distribution of proportion.

c) (4 points) In a random sample of 600 18- to 29-year-olds, what is the probability that more than 75% would prefer to start their own business?

Answer 1

Solution:

Given ,

p = 70% = 0.70(population proportion)

n = 600 (sample size)

Let be the sample proportion.

a)

mean of this sampling distribution of proportion

=

= p

= 0.70

b)

standard deviation of this sampling distribution of proportion

=

=

=

=  0.01870828693

c)

P[more than 75%]

= P[ > 0.75]

=  

=  P(Z > (0.75-0.70)/0.01870828693)

= P(Z > 2.67)

=1 - P(Z < 2.67)

= 1 - 0.9962 ...use z table

= 0.0038