Chapter 6.4, Problem 17E

Mistakes in Restaurant Bills About 12.5% of restaurant bills are incorrect. If 200 bills are selected at random, find the probability that at least 22 will contain an error. Is this likely or unlikely to occur?

To determine

To find: The probability that at least 22 bills will contain an error.

To identify: Whether the event is unlikely to occur.

Answer to Problem 17E

The probability that at least 22 bills will contain an error is 0.7734.

The event that at least 22 bills will contain an error is likely to occur.

Explanation of Solution

Given info:

The random variable X denote the mistakes of restaurant bills and it follows the binomial distribution with parameters $\left(n=200,p=0.125\right).$

Calculation:

Let n be the random sample of restaurant bills and p be the probability that restaurant bills are incorrect. That is, $n=200\text{, }p=\mathrm{0.125.}$

Check the conditions for normal approximation:

Condition 1:

$\begin{array}{c}np\ge 5\\ 200\left(0.125\right)\ge 5\\ 25\ge 5\end{array}$

Condition 2:

$\begin{array}{c}n\left(1-p\right)\ge 5\\ 200\left(1-0.125\right)\ge 5\\ 175\ge 5\end{array}$

Hence, both the conditions are satisfied.

The mean of the random variable (X) is,

$\begin{array}{c}\mu =np\\ =200\left(0.125\right)\\ =25\end{array}$

The standard deviation of the random variable (X) is,

$\begin{array}{c}\sigma =\sqrt{np\left(1-p\right)}\\ =\sqrt{200\left(0.125\right)\left(1-0.125\right)}\\ =4.68\end{array}$

The probability that at least 22 bills will contain an error is,

$\begin{array}{l}P(X \ge 22)=P\left(X>22-0.5\right)\left(\text{Using continuity correction}\right)\\ =P\left(\frac{X-\mu }{\sigma }>\frac{21.5-25}{4.68}\right)\\ =P\left(Z>-0.75\right)\end{array}$

The z score is obtained as follows:

For $P\left(Z>-0.75\right)$

Excel procedure:

Step by step procedure to obtain z score using the Excel:

• Select a cell in the Excel
• Type “ $=NORMSDIST\left(-0.75\right)$”
• Press Enter.

It returns the probability value is 0.2266.

Solving the below equation gives the required probability,

$\begin{array}{c}P\left(Z>-0.75\right)=1-P\left(Z\le -0.75\right)\\ =1-0.2266\\ =0.7734\end{array}$

Conclusion:

Hence, the probability that at least 22 bills will contain an error is, 0.7734.

Justification:

Unlikely events:

If the probability for any event is less than or equal to 0.05, then it is considered to be unlikely event.

Here, the probability that at least 22 bills will contain an error is 0.7734 which is not less than 0.05.

Thus, the event is likely to occur.