Data on consumer spending per capita or industrial purchasing trends would be identified in the ________ section of a global mar
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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:
Now the probability is given as
The value of P(Z<1) is calculated from the table which is given as
0.84134
So the equation becomes
So the probability of having a weekly profit of more than 2000 is 0.1587 or 15.87%.