Question

According to the U.S. Census Bureau, 20% of the workers in Atlanta use public transportation. Suppose 25 Atlanta workers are randomly selected. (Hint: use sampling distribution) (a) What is the standard deviation of the sample proportion of the selected workers who use public transportation? (5 points) (a) What is the probability that the proportion of the selected workers who use public transportation is less than 32%? (5 points) (b) What is the probability that the proportion of the selected workers who use public transportation is greater than 48%? (5 points)

Answer 1

Let X be the no. of workers in Atlanta that use public transportation.
X~Binomial(n,p)

A random sample of 25 workers is selected therefore
X~Binomial(25,0.2)

a) The standard deviation is given as:

b)

32% people is equal to 25*0.32=8
Therefore we need to calculate the probability

Where:

P(X=0)	0.000792
P(X=1)	0.006338
P(X=2)	0.024561
P(X=3)	0.061402
P(X=4)	0.111291
P(X=5)	0.155807
P(X=6)	0.175283
P(X=7)	0.162763
Total	0.698237

P(X<8)=0.698

b)

48% people is equal to 25*0.48=12
Therefore we need to calculate the probability

P(X=0)	0.000792
P(X=1)	0.006338
P(X=2)	0.024561
P(X=3)	0.061402
P(X=4)	0.111291
P(X=5)	0.155807
P(X=6)	0.175283
P(X=7)	0.162763
P(X=8)	0.127158
P(X=9)	0.084772
P(X=10)	0.048744
P(X=11)	0.024372
Total	0.983284

P(X>12)=1-0.9832=0.0168