Problem 184SE

A city commissioner claims that 80% of the people living in the city favor garbage collection by contract to a private company over collection by city employees. To test the commissioner’s claim, 25 city residents are randomly selected, yielding 22 who prefer contracting to a private company.

a If the commissioner’s claim is correct, what is the probability that the sample would contain at least 22 who prefer contracting to a private company?

b If the commissioner’s claim is correct, what is the probability that exactly 22 would prefer contracting to a private company?

c Based on observing 22 in a sample of size 25 who prefer contracting to a private company, what do you conclude about the commissioner’s claim that 80% of city residents prefer contracting to a private company?

Solution :

Step 1 of 3:

Let p denotes the 80% of the people living in the city favour garbage collection by contract to a private company.

So p=80%=0.80.

The 25 residents are selected randomly.

So n=25.

Then the number of 22 people who prefer contracting to a private company.

Our goal is:

a). If the commissioner’s claim is correct, we need to find the probability that the sample

would contain at least 22 who prefer to contracting a private company.

b). We need to find the probability that exactly 22 would prefer contracting to a private company.

c). Based on observing 22 in a sample of size 25 who prefer contracting to a private

company, we need to conclude about the commissioner's claim that 80% of city

residents prefer contracting to a private company.

a).

If the commissioner’s claim is correct.

We have to find the probability that the sample would contain at least 22 who prefer to contracting a private company.

The formula for the binomial probability is

P(X=a)=

Then the probability of at least 22 is

P(X22)=P(X=22)+P(X=23)+P(X=24)+P(X=25)

We know that n=25, p=0.80.

Where, P(X=22),P(X=23),P(X=24),P(X=25) is obtained from Excel by using the function “=Binomdist(x,n,p,false)”

Then the table is given below.

X |
p(x 22) |

22 |
0.135768036 |

23 |
0.070835497 |

24 |
0.023611832 |

25 |
0.003777893 |

Total |
0.233993259 |

Now,

P(X22)= P(X=22)+P(X=23)+P(X=24)+P(X=25)

P(X22)= 0.23399

Therefore the probability of at least 22 is 0.23399.