Problem 21E

In a recent study 90 percent of the homes in the United States were found to have large- screen TVs. In a sample of nine homes, what is the probability that:

a. All nine have large-screen TVs?

b. Less than five have large-screen TVs?

c. More than five have large-screen TVs?

d. At least seven homes have large-screen TVs?

Solution 21E

Step1 of 5:

Let us consider a random variable ‘X’ it presents the number of homes in the United States were found to have large- screen TV’s. Also, we have ‘n = 9 and ’.

Here our goal is:

a). We need to find the probability that all nine have large-screen TV’s.

b). We need to find the probability that less than five have large-screen TV’s.

c). We need to find the probability that more than five have large-screen TV’s.

d). We need to find the probability at least seven homes have large-screen TV’s.

Step2 of 5:

a).

We know that ‘x’ follows a binomial distribution with parameters ‘n = 9 and ’. Then, the probability mass function of the binomial distribution is:

Now,

Therefore, P(X = 9) = 0.3875.

Step3 of 5:

b).

The probability that less than five have large-screen TV’s is:

Where, is obtained from Excel by using the function “=Binomdist(X,n,false)”

X |
P(X < 5) |

1 |
8.1E-08 |

2 |
2.92E-06 |

3 |
6.12E-05 |

4 |
0.000827 |

Total |
0.000891 |

Therefore, P(X < 5) = 0.000891.

Step4 of 5:

c).

The probability that more than five have large-screen TV’s is:

Where,

Where, is obtained from Excel by using the function “=Binomdist(X,n,false)”

X |
P(X5) |

1 |
8.1E-08 |

2 |
2.92E-06 |

3 |
6.12E-05 |

4 |
0.000827 |

5 |
0.00744 |

Total |
0.008331 |

Therefore, P(X5) = 0.008331.

Now,

Therefore,

Step5 of 5:

d).

The probability at least seven homes have large-screen TV’s is:

Where,

Where, is obtained from Excel by using the function “=Binomdist(X,n,false)”

X |
P(X < 7) |

1 |
8.1E-08 |

2 |
2.92E-06 |

3 |
6.12E-05 |

4 |
0.000827 |

5 |
0.00744 |

6 |
0.044641 |

Total |
0.0530 |

Therefore, P(X < 5) = 0.0530.

Now,

Therefore,