Chapter 5.3, Problem 1AC

Unsanitary Restaurants

Health officials routinely check the sanitary condition of restaurants. Assume you visit a popular tourist spot and read in the newspaper that in 3 out of every 7 restaurants checked, unsatisfactory health conditions were found. Assuming you are planning to eat out 10 times while you are there on vacation, answer the following questions.

1. How likely is it that you will eat at three restaurants with unsanitary conditions?

2. How likely is it that you will eat at four or five restaurants with unsanitary conditions?

3. Explain how you would compute the probability of eating in at least one restaurant with unsanitary conditions. Could you use the complement to solve this problem?

4. What is the most likely number to occur in this experiment?

5. How variable will the data be around the most likely number?

6. How do you know that this is a binomial distribution?

7. If it is a binomial distribution, does that mean that the likelihood of a success is always 50% since there are only two possible outcomes? Check your answers by using the following computer-generated table.

Mean = 4.29 Std. dev. = 1.56492

X	P(X)	Cum. prob.
0	0.00371	0.00371
1	0.02784	0.03155
2	0.09396	0.12552
3	0.18793	0.31344
4	0.24665	0.56009
5	0.22199	0.78208
6	0.13874	0.92082
7	0.05946	0.98028
8	0.01672	0.99700
9	0.00279	0.99979
10	0.00021	1.00000

To determine

The probability that the person is likely to eat at all three restaurants with unsanitary conditions.

Answer to Problem 1AC

The probability the person is likely to eat at all three restaurants with unsanitary conditions is 0.186.

Explanation of Solution

Given info:

A person is planning to eat outside 10 times during a vacation. Health officials checks the sanitary condition of restaurant and found that every 3 out of 7 restaurants checked was having an unsatisfactory health conditions.

Calculation:

Define the random variable x as the number of restaurants with unsanitary conditions. Here, the total number of restaurants are (n) is 10 and each restaurant is independent from other restaurants. Also, there are two possible outcomes (having an unsanitary conditions or didn’t have an unsanitary conditions) and the probability of that the selected restaurant having an unsanitary conditions gives the probability (p) $0.43\left(=\frac{3}{7}\right)$. Thus, x follows binomial distribution.

The binomial distribution formula is,

$P\left(X\right)=\frac{n!}{\left(n-X\right)!X!}{p}^X{q}^{n-X}$

Where, n is the number of trials, x is the number of successes among n trials, p is the probability of success and q is the probability of failure.

Substitute n as 10, p as 0.43, q as $0.57\left(=1-0.43\right)$

$\begin{array}{c}P\left(3\right)=\frac{10!}{\left(10-3\right)!3!}{\left(0.43\right)}^3{\left(0.57\right)}^{10-3}\\ =120\left(0.0795\right)\left(0.0195\right)\\ =0.186\end{array}$

Thus, the probability the person is likely to eat at all three restaurants with unsanitary conditions is 0.186.

To determine

The probability that the person is likely to eat at four or five restaurants with unsanitary conditions.

Answer to Problem 1AC

The probability the person is likely to eat at four or five restaurants with unsanitary conditions is 0.469.

Explanation of Solution

Here, “Eating at four or five restaurants with unsanitary conditions gives the values of X as 4 and 5.

Substitute n as 10, p as 0.43, q as $0.57\left(=1-0.43\right)$

$\begin{array}{c}P\left(4\right)+P\left(5\right)=\left[\frac{10!}{\left(10-4\right)!4!}{\left(0.43\right)}^4{\left(0.57\right)}^{10-4}\right]+\left[\frac{10!}{\left(10-5\right)!5!}{\left(0.43\right)}^5{\left(0.57\right)}^{10-5}\right]\\ =210\left(0.0342\right)\left(0.0343\right)+252\left(0.0147\right)\left(0.0601\right)\\ =0.246+0.223\\ =0.469\end{array}$

Thus, the probability the person is likely to eat at four or five restaurants with unsanitary conditions is 0.469.

To determine

To explain: The way to compute the probability of eating in at least one restaurant with unsanitary conditions.

To find: The probability of eating in at least one restaurant with unsanitary conditions.

Answer to Problem 1AC

The way to compute the probability of eating in at least one restaurant with unsanitary conditions is given below:

Eating in at least one restaurant is same as eating in 1 or more than 1 restaurants.

$P\left(\text{Atleast one restaurant}\right)=1-P\left(\text{Eating in none of the restaurants}\right)$

The probability of eating in at least one restaurant with unsanitary conditions is 0.9964.

Explanation of Solution

Calculation:

Substitute n as 10, p as 0.43, q as $0.57\left(=1-0.43\right)$ and x as 0.

$\begin{array}{c}P\left(\text{Atleast one restaurant}\right)=1-P\left(\text{Eating in none of the restaurants}\right)\\ =1-P\left(0\right)\\ =1-\left[\frac{10!}{\left(10-0\right)!0!}{\left(0.43\right)}^0{\left(0.57\right)}^{10-0}\right]\\ =1-0.0036\end{array}$

                                           =0.9964

Thus, the probability of eating in at least one restaurant is 0.9964.

To determine

The most likely number that could occur in the given experiment.

Answer to Problem 1AC

The most likely number that could occur in the given experiment is the event of eating in 4 restaurants with unsanitary conditions.

Explanation of Solution

Calculation:

Software procedure:

Software procedure for calculating the probability is given below:

• Choose Calc > Probability Distributions > Binomial Distribution.
• Choose Probability.
• Enter Number of trials as 10 and Event probability as 0.43.
• In Input columns, enter the column containing the values 0, 1, 2, 3,…10.
• Click OK.

Output obtained from MINITAB is given below:

Justification:

From the MINITAB output it can be observed that the probability of eating in 4 restaurants has the highest probability of 0.2462.

The probability for 4 restaurants is the highest, but the expected number of restaurants that a person would eat is $\left(10\right)\left(0.43\right)=4.29$.

To determine

To find: The variability of the data around the most likely number.

Answer to Problem 1AC

The variability of the data around the most likely number would be 1.565 restaurants.

Explanation of Solution

Calculation:

Standard deviation:

$\sigma =\sqrt{npq}$

Substitute n as 10, p as 0.43 and q as $0.57\left(1-0.43\right)$

$\begin{array}{c}\sigma =\sqrt{\left(10\right)\left(0.43\right)\left(0.57\right)}\\ =\sqrt{2.451}\\ \approx 1.565\end{array}$

Thus, the variability of the data around the most likely number would be 1.565 restaurants.

To determine

To justify: The reason for identifying the given experiment as binomial.

Answer to Problem 1AC

The given experiment satisfies all the “requirements of binomial distribution”.

Explanation of Solution

Justification:

Requirements of binomial distribution:

• There will be a fixed number of trials.
• There are only two possible outcomes (success and failure).
• The probability of success remains constant.
• The outcomes obtained from each trial are independent of one another.

Here, the number of restaurants is 10 and it gives the fixed number of trials, there are only two possible outcomes either a restaurant might have unsanitary condition or a restaurant might not have unsanitary condition. The probability of that the selected restaurant having an unsanitary conditions gives the probability (p) $0.43\left(=\frac{3}{7}\right)$. Also, the outcomes are independent from each other.

Thus, the given experiment is a binomial distribution.

To determine

To check: Whether the mean of the likelihood of success is always 50%.

Answer to Problem 1AC

The mean of the likelihood of success will vary from situation to situation.

Explanation of Solution

Given info:

Use the computer generated table for checking.

Justification:

In binomial distribution the probability of success remains constant and it is 50% but it varies for situation to situation. Just having two outcomes will not assurance for equal probabilities of success.