In: Statistics and Probability
QUESTION 2
According to Michael Theatre Ltd., the mean cost to run a nightly theatre performance is $3,200 with a standard deviation of $460. Performance costs are known to follow a normal probability distribution.
Required:
I Need the full calculation and explanation of the answers.
We are given that the mean performance cost =$3200 and =$460. We know that any Normal distribution can be transformed to a standard normal variate by the transformation
a. Percentage of the performances will cost less than $2,850 to run =
Percentage of the performances will cost less than $2,850 to run
We can either look into the table(Fisher & Yates) or use the EXCEL function.NORM.S.DIST(-0.7609,TRUE).
. ie 22.34% of the performances will cost less than $2850.
b.Percentage of the performances will cost between $3,250 and $3,520 to run
For $3250,
For $3520,
Percentage of the performances will cost between $3,250 and $3,520 to run
This can be calculated by
Percentage of the performances will cost between $3,250 and $3,520 to run=0.2131 or 21.31%
c. Percentage of the performances will cost between $2,300 and $3,520 to run
For $2300,
For $3520,
Percentage of the performances will cost between $2,300 and $3,520 to run
Percentage of the performances will cost between $2,300 and $3,520 to run =0.7315 or 73.15%.
d.Percentage of the performances will cost less than $3,750 to run
Percentage of the performances will cost less than $3,750 to run
or 88.41%
Percentage of the performances will cost less than $3,750 to run=88.41%
e.Six percent (6 %) of all shows will cost less than what dollar amount. here we need to first find out that Z for which
This will be obtained by the Inverse Normal function to find the ordinate. NORM.S.INV(0.06) in Excel or the Table 1 of the Fisher & Yates. which is -1.5548
Six percent (6 %) of all shows will cost less than $2485.