Tipping. In the article "Are Christian/Religious Pcople Poor Tippers?" (hournol of A,piled Sockal Psychology, Vol. 43. Issue 5 . pp. 928-935), M. Lynn and B. Katz report that, for customers who receive bad service, the mean percentage tip is $8.56\%$of the bill with a standard deviation of $5.37\%$. Let $x$represent the percentage of the bill that a customer tips when he or she receives bad service.

a. What percentage of tips for bad service are at least 0 se?

b. Assuming that $x$ is aprroximalely noemally distribuled, uve normalcurve areas to determine the approximate percentage of tips for bad service that are at least $0\%$

c. Based on your results from parts (a) and (b), do you think that the percentage of the bill that a customer tips when he or she receives bad service is appeexinately a normally distributed variable? Explain your answer.

To determine the fraction of bad service tips that are at least$0\%$

The bill's percentage tip to the unsatisfactory service The mean of $\left(x\right)$is$\mu =8.56\%$and the standard deviation is$\sigma =5.37\%$

Negative suggestions are not possible based on the facts provided. As a result, bad-service tips are always more than or equal to $0\%$. As a result, the probability of receiving a tip for poor service is at least $0\%$is $100\%$

To discover the percentage of tips for terrible service that are at least $0\%$ if $X$ is roughly followed by a normal distribution.

The random variable $x$'s $z$-score is calculated as follows:

$z=\frac{x-\mu }{\sigma }$

Calculate the $z$- scores:

localid="1651723581461" $$\begin{gathered} z=\frac{0-8.56}{5.37} \\ \approx -1.59 \end{gathered}$$

Using the normal probability table in Appendix A, calculate the corresponding probability:

$P(x>0)=P(z>-1.59)$

$=1-0.0559$

$=94.41\%$

To determine whether $x$follows a normal distribution roughly based on the results of parts (a) and (b) (b).

The percentages in parts (a) and (b) above should be equal to $100\%$, yet the outcomes of part (a) and (b) differ significantly (b). As a result, it is impossible to establish that $x$ follows a normal distribution.