On 100 different days, a traffic engineer counts the number of cars that pass through a certain intersection between 5 p.m. and 5:05 p.m. The results are presented in the following table.

Number of Cars |
Number of Days |
Proportion of Days |

0 |
36 |
0.36 |

1 |
28 |
0.28 |

2 |
15 |
0.15 |

3 |
10 |
0.10 |

4 |
7 |
0.07 |

5 |
4 |
0.04 |

a. Let X be the number of cars passing through the intersection between 5 P.M. and 5:05 P.M. on a randomly chosen day. Someone suggests that for any positive integer x, the probability mass function of X is p1(x) = (0.2) (0.8) x. Using this function, compute P(X = x) for values of x from 0 through 5 inclusive.

b. Someone else suggests that for any positive integer x, the probability mass function is p2(x) = (0.4)(0.6)x. Using this function, compute P(X = x) for values of x from 0 through 5 inclusive.

c. Compare the results of parts (a) and (b) to the data in the table. Which probability mass function appears to be the better model? Explain.

d. Someone says that neither of the functions is a good model since neither one agrees with the data exactly. Is this right? Explain.

Answer :

Step 1 of 5 :

Given, a traffic engineer counts the number of cars that pass through a certain intersection between 5.pm and 5:05 pm. On 100 different days

Number of cars |
Number of days |
Proportion of days |

0 |
36 |
0.36 |

1 |
28 |
0.28 |

2 |
15 |
0.15 |

3 |
10 |
0.10 |

4 |
7 |
0.07 |

5 |
4 |
0.04 |

Step 2 of 5 :

Where X be the number of cars passing through the intersection between 5 P.M. and 5:05 P.M.For any positive integer x, the probability mass function of X is

(x) = (0.2) (0.8.

The claim is to compute P(X = x) for values of x from 0 through 5 inclusive

Then, (0) = (0.2) (0.8.

= 0.2

(1) = (0.2) (0.8.

= 0.16

(2) = (0.2) (0.8.

= 0.128

(3) = (0.2) (0.8.

= 0.1024

(4) = (0.2) (0.8.

= 0.0819

(5) = (0.2) (0.8.

= 0.0655

Step 3 of 5 :

b)

For any positive integer x, the probability mass function of X is

(x) = (0.4) (0.6.

The claim is to compute P(X = x) for values of x from 0 through 5 inclusive

Then, (0) = (0.4) (0.6.

= 0.4

(1) = (0.4) (0.6.

= 0.24

(2) = (0.4) (0.6.

= 0.144

(3) = (0.4) (0.6.

= 0.0864

(4) = (0.4) (0.6.

= 0.0518

(5) = (0.4) (0.6.

= 0.0311