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Weekly demand for tennis balls at The Racquet Club is normally distributed , with a mean of 35 cases and a standard deviation of 5 cases . The club gets a profit of $ 50 per case (a ) Simulate 52 weeks of demand and calculate the average weekly profit. Make all demand values integers in your model . (b ) What is the probability that weekly profit will be \$2,00 or more ?

Sagot :

danialamin

Answer:

a-The average weekly profit is $1767.31

b- The probability of having a weekly profit of more than 2000 is 0.1587 or 15.87%.

Explanation:

a

The weekly average profit for the simulation is given where first the values are simulated using R which is given as below:

x<-round(rnorm(n,m,s))

Here

  • round converts all the values of the simlation to integer.
  • rnorm is the command for simulation
  • n is the number of values which is 52 in this case
  • m is the mean of the values which is 35
  • s is the value of standard deviation which is 5 cases.

The values of x are as follows

[1] 36 49 30 29 34 36 32 28 32 29 32 27 40 32 30 37 43 30 42 30 31 34 36 38 28 29 32 42 36 35

[31] 37 41 34 39 37 46 34 44 45 41 41 29 36 38 35 32 36 39 30 38 40 27

Now using these values, the average of the simulation values is cacluated as follows:

mean(x)

35.3462

Now using this with the value of profit of $50 gives:

Average Profit=$50 x 35.3462

Average Profit=$1767.31

The average weekly profit is $1767.31

b-

First number of cases are required so that the value will be greater than 2000 it is given as

Number of cases=2000/50=40

So firstly the Z-score is calculated which is as below:

[tex]Z=\dfrac{x-\mu}{\sigma}\\Z=\dfrac{40-35}{5}\\Z=1[/tex]

Now the probability is given as

[tex]P(X\geq 40)=P(Z\geq 1)\\P(X\geq 40)=1-P(Z< 1)[/tex]

The value of P(Z<1) is calculated from the table which is given as

0.84134

So the equation becomes

[tex]P(X\geq 40)=1-P(Z< 1)\\P(X\geq 40)=1-0.8413\\P(X\geq 40)=0.1587[/tex]

So the probability of having a weekly profit of more than 2000 is 0.1587 or 15.87%.

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