Chapter 16, Problem 45E

(a)

To determine

To find: the probability that the researcher sells above 2000 cups of coffee in one week.

Answer to Problem 45E

1.63

Explanation of Solution

Given:

Cups of coffee

mean = 320 cups

standard deviation = 20 cups

Doughnut

mean = 150 doughnut

standard deviation = 12 doughnuts

Formula used:

  $Z-\text{score = }\frac{x-\mu }{\sigma }$

Calculation:

To find the total coffee cups per week, the averages of 320 cups per day must be multiplied by 6 days in one week, when the shop is open.

  $\begin{array}{c}\text{weekly mean = Daily mean }\times \text{ Days open}\\ \text{=320}\times \text{6}\\ \text{=1,920 cups}\end{array}$

To evaluate the standard deviation in cups within a week, the difference of doughnuts each week needs to be calculated by applying the square root of the standard deviatin for the six days that the shop has been opened.

  $\begin{array}{c}\mathrm{var}\left(\text{weekly Doughnuts}\right){\text{ = Daily Standard deviation}}^2\times \text{ Days open}\\ {\text{=20}}^2\times \text{6}\\ \text{=2400}\end{array}$

For the standard deviation

  $\begin{array}{c}\text{weekly Standard Deviation=}\sqrt{\mathrm{var}\left(\text{weekly Doughnuts}\right)}\\ \text{=}\sqrt{2400}\\ =48.99\text{cups of coffee}\end{array}$

For the z score

  $\begin{array}{c}Z-\text{score = }\frac{x-\mu }{\sigma }\\ =\frac{2000-1920}{48.99}\\ =1.63\end{array}$

The table reveals that the 1.63 z-score is 0.9483 or 94.84% of the probability that fewer than 2,000 cups will go to the store. Thus, 0.0516 or 5.16 per cent (1 minus 94.84 per cent) is required to sell the shop over 2,000 cups of coffee a week

(b)

To determine

To Explain: that researcher can reasonably predict to has one day’s profit of above $300.

Explanation of Solution

Given:

Cups of coffee

mean = 320 cups

standard deviation = 20 cups

Doughnut

mean = 150 doughnut

standard deviation = 12 doughnuts

Calculation:

It is considered that 50 cents for each cup of coffee and 40 cents for every doughnut make a profit from the market. it has to decide if the shop is going to receive more than $300 a day. Second, as the number of coffee cup (c) and doughnuts (D) sales, it may convey the profit of the company.

  $\text{profit = \$0}\text{.50}\times \text{C+\$0}\text{.40}\times \text{D}$

Now, estimate the mean

  $\begin{array}{c}\text{profit, mean = \$0}\text{.50}\times \text{C+\$0}\text{.40}\times \text{D}\\ \text{=\$0}\text{.50}\times \text{320+\$0}\text{.40}\times \text{150}\\ \text{=\$220}\end{array}$

Therefore, the standard deviation of profits must be measured using the profits equation and the variance squaring value per unit and the standard deviation between the two goods must be determined.

  $\begin{array}{c}\mathrm{var}\left(\text{profit}\right)\text{= \$0}{\text{.50}}^2\times \mathrm{var}{\left(\text{C}\right)}^2\text{+\$0}\text{.40}\times \mathrm{var}{\left(\text{D}\right)}^2\\ \text{=\$0}{\text{.50}}^2\times {\left(\text{20}\right)}^2\text{+\$0}\text{.40}\times \text{var}{\left(12\right)}^2\\ \text{=123}\text{.04}\end{array}$

For the standard deviation

  $\begin{array}{c}\text{profit Standard Deviation=}\sqrt{\mathrm{var}\left(\text{profit}\right)}\\ \text{=}\sqrt{123.04}\\ =\$11.09\end{array}$

The estimated daily benefit is thus $220 and the standard deviation is $11.09.it would then infer that a day's profit of over $300 cannot be expected for the company, since there will be over seven standard deviations away from the average of $220.

(c)

To determine

To find: the probability that on any mention day researcher would sell one doughnut to greater than half of the coffee customers.

Explanation of Solution

Given:

Cups of coffee

mean = 320 cups

standard deviation = 20 cups

Doughnut

mean = 150 doughnut

standard deviation = 12 doughnuts

Calculation:

It was important to assess the probability that more than half of its coffee customers would buy a doughnut on every day store. The doughnut customers (D) and half of coffee clients (C) have to be subtracted in order to demonstrate the gap.

Difference = D- 0.5(C).

  $\begin{array}{c}\text{Difference, mean=D-0}\text{.5C}\\ \text{=150-0}\text{.5}\times \text{320}\\ \text{=-10}\end{array}$

The difference must then be evaluated by means of the difference equation and the variation shall be estimated according to the standard deviation of the doughnut as well as of the coffee consumer. The difference between the two variables must then be determined by the total of the variances.

  $\begin{array}{c}\mathrm{var}\left(\text{Difference}\right)\text{=var}{\left(\text{D}\right)}^2\text{+0}{\text{.5}}^2\mathrm{var}\left(\text{C}\right)\\ =\mathrm{var}{\left(12\right)}^2+{0.5}^2\times \mathrm{var}{\left(20\right)}^2\\ =244\end{array}$

  $\begin{array}{c}\text{Difference, Standard Deviation=}\sqrt{\mathrm{var}\left(\text{Difference}\right)}\\ \text{=}\sqrt{244}\\ =15.62\end{array}$