Q.62 - The Practice of Statistics for AP

Question

How many people are in a car? A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban interchange. Suppose that this count has a mean of $1.5$. and a standard deviation of $0.75$ in the population of all cars that enter this interchange during rush hours.

(a)  Could the exact distribution of the count be Normal? Why or why not? 

(b) Traffic engineers estimate that the capacity of the interchange is $700$ cars per hour. Find the probability that $700$ cars will carry more than $1075$ people. Show your work. (Hint: Restate this event in terms of the mean number of people x per car.) 

Step-by-Step Solution

Answer

From the given information,

a) No, the exact distribution of the count is not Normal.

b) The probability is $1.038$

Part (a) Step 1: Given Information

It is given in the question that, the population mean $\left(\mu \right)=1.5$

population standard deviation $\left(\sigma \right)=0.75$

Could the exact distribution of the count be Normal? Why or why not?  

Part (a) Step 2: Explanation

It is known that the normal distribution is entitled to take any real number. Here, the count is taking only values that are positive integers. Thus, the distribution of count is not normal.

Part (b) Step 1: Given Information

It is given in the question that, the population mean $\left(\mu \right)=1.5$ 

population standard deviation $\left(\sigma \right)=0.75$

Find the probability that $700$ cars will carry more than $1075$ people.

Part (b) Step 2: Explanation

There are total $700$ passengers. The average number of passengers can be calculated as:

$\stackrel{-}{X}=\frac{1075}{700}$

$=1.536$

The probability that more than $1075$ people would be carried out by the $700$ cars is calculated as follows:

$P(\overline{X}>21.536)=P(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{1.536-\mu }{\frac{\sigma }{\sqrt{n}}})$

                         $=P(Z>\frac{1.536-1.5}{\frac{0.75}{\sqrt{100}}})$

                        $=P(Z>1.26)\text{(From standard normal table)}$

                        $=0.1038$

Thus, the required probability is $1.038$