At a traveling carnival, a popular game is called the “Cash Grab.” In this game, participants step into a sealed booth, a powerful fan turns on, and dollar bills are dropped from the ceiling. A customer has 30 seconds to grab as much cash as possible while the dollar bills swirl around. Over time, the operators of the game have determined that the mean amount grabbed is \(13 with a standard deviation of \)9. They charge \(15 to play the game and expect to have 40 customers at their next carnival.

a. What is the probability that an SRS of 40 customers grab an average of \)15 or more?

b. How much should the operators charge if they want to be 95% certain that the mean amount grabbed by an SRS of 40 customers is less than what they charge to play the game?

The mean is $\mathrm{\mu }=13$and standard deviation is $\mathrm{\sigma }=9$

The number of customers

The sample mean $\overline{\mathrm{x}}=15$

The following concept was used:

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}$

The sampling distribution of the sample mean $\overline{\mathrm{x}}$is also normal because the population distribution is normal.

Z-score is $\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}=\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{{\displaystyle \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}}}=\frac{15-13}{{\displaystyle \frac{9}{\sqrt{40}}}}=1.41$

Using the normal probability, the associating probability is calculated.

$\mathrm{P}\left(\mathrm{Z}<1.41\right)$is typical normal probability table in the appendix has a row beginning with 1.4 and a column beginning with .01.

$\begin{array}{c}\mathrm{P}\left(\overline{\mathrm{X}}\ge 15\right)=\mathrm{P}\left(\mathrm{Z}>1.41\right)\\ =1-\mathrm{P}\left(\mathrm{Z}<1.41\right)\\ =1-0.9207\\ =0.0793\\ =7.93\text{\%}\end{array}$

The mean is $\mathrm{\mu }=13$and standard deviation is$\mathrm{\sigma }=9$

The number of customers $\mathrm{n}=40$

$\mathrm{P}\left(\overline{\mathrm{X}}\le \overline{\mathrm{x}}\right)=95\mathrm{\%}$

The following concept was used:

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}$

Determine the z-score that corresponds to a probability of 95 percent or 0.95 in the normal probability table and The probability 0.95 lies exactly between 0.9495 and 0.9505, where the z-scores 1.64 and 1.64 correspond to the probability 0.95, which is 1.645.

$\mathrm{z}=1.645$

Z-score is $\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}=\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{{\displaystyle \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}}}=\frac{15-13}{{\displaystyle \frac{9}{\sqrt{40}}}}=1.41$

The z-two score's found expressions must then be equal:

$\frac{\overline{x}-13}{\frac{9}{\sqrt{40}}}=1.645$

Each side should be multiplied by a $\frac{9}{\sqrt{40}}$

$\overline{\mathrm{x}}-13=1.645\left(\frac{9}{\sqrt{40}}\right)$

To each side, add 13:

$\overline{\mathrm{x}}+13=1.645\left(\frac{9}{\sqrt{40}}\right)$

Determine:

$\mathrm{x}=15.34$

As a result, the business should charge is 15.34