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?
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
Get Answers For Free
Most questions answered within 1 hours.