DValuTown, a small midwestern grocery store, is assessing its shelf inventory of
ID: 3020310 • Letter: D
Question
DValuTown, a small midwestern grocery store, is assessing its shelf inventory of cans of green beans. It needs some help understanding its demand patterns to better determine an appropriate target shelf level of inventory. The firm presently has weekly demand data for this stock keeping unit over the past 500 weeks. It has found that the average weekly demand is 100 cans with a stand deviation of 6 cans. In addition, the average weekly demand has been found to be normally distributed. Listed below you will find demand information for 20 randomly sampled weekly demand values.
1. Determine the mean demand of these 20 sample observations
2006/20= 100.3
2. Using the population standard deviation, how many standard deviations away from the population mean is the smallest observation in the sample data above?
1.5 ?
3. If you were to draw a random sample of 20 observations from the entire population of 500 demand observations, what is the likelihood (probability) that you would find a larger sample mean than the value you calculated in question 1?
4. ValuTown has conducted some analysis that reveals that approximately 5% of the weekly demands have an average value below 88 cans. In a randomly drawn sample of 200 weekly demand observations, what is the likelihood the sampled mean weekly demand would be below 86 cans?
5. Based upon the population data, what percentage of demand values would be expected to exceed 115 cans?
6. Based upon the population data, what is the demand value that you would expect to be exceeded only 25 percent of the time?
7. Based upon the population data, what is the demand that you would expect to exceed 90 percent of the time?
8. Based upon the population data, what is the likelihood of weekly demand being between 85 and 115 cans?
9. ValuTown has conducted some analysis that reveals the average customer buys 3 cans of green beans per transaction. How many transactions does ValuTown process on average per week.
Observation Demand 1 100 Cans 2 105 3 112 4 97 5 92 6 114 7 98 8 91 9 101 10 94 11 101 12 97 13 95 14 102 15 91 16 110 17 103 18 94 19 102 20 107Explanation / Answer
DValuTown, a small midwestern grocery store, is assessing its shelf inventory of cans of green beans.
It needs some help understanding its demand patterns to better determine an appropriate target shelf level of inventory.
The firm presently has weekly demand data for this stock keeping unit over the past 500 weeks.
Let X be the weekly demand.
Average weekly demand (mu) = 100 cans
standard deviation (sd) = 6 cans
Listed below you will find demand information for 20 randomly sampled weekly demand values.
In addition, the average weekly demand has been found to be normally distributed.
Determine the mean demand of these 20 sample observations
sample mean of weekly demand (Xbar) = sum of demands / total number of demands
Xbar = 2006 / 20 = 100.3
2. Using the population standard deviation, how many standard deviations away from the population mean is the smallest observation in the sample data above?
smallest obsrevation of the data is 91.
We have to calculate how many standard deviations away from the population mean is the smallest observation in the sample data above.
That is we are given mu - k*sd = 91
100 - k * 6 = 91
100 - 91 = k * 6
9 = k * 6
k = 9 / 6 = 1.5
Therefore, 1.5 standard deviations away from the population mean is the smallest observation in the sample data above.
3. If you were to draw a random sample of 20 observations from the entire population of 500 demand observations, what is the likelihood (probability) that you would find a larger sample mean than the value you calculated in question 1?
population mean = 100
sample mean of demand (Xbar) = 100.3
Now we have to calculate P(Xbar > 100.3)
By using central limit theorem,
Xbar ~ Normal(mu = 100, sd = 6)
convert Xbar into z-score,
z = (Xbar - mu) / (sd/sqrt(n))
Xbar = 100.3
n = 20
z = (100.3 - 100) / (6 / sqrt(20))
z = 0.2236
Now we have to find P(Z > 0.2236)
EXCEL syntax :
= 1 - NORMSDIST(z) (EXCEL always gives left tail area)
where z = 0.2236
P(Z > 0.2236) = 0.4115
4. ValuTown has conducted some analysis that reveals that approximately 5% of the weekly demands have an average value below 88 cans. In a randomly drawn sample of 200 weekly demand observations, what is the likelihood the sampled mean weekly demand would be below 86 cans?
That is we are given that,
sample size (n) = 200
We have to find P(Xbar < 86).
z-score for Xbar = 86 is,
z = (86 - 100) / (6/sqrt(200))
z = -32.9983
Now we have to find P(Z < -32.9983)
EXCEL syntax is,
=NORMSDIST(z)
where z is test statistic value.
P(Z < -32.9983) = 4.2934*10-239
which is approximately 0.
5. Based upon the population data, what percentage of demand values would be expected to exceed 115 cans?
We have to find P(X > 115)
Convert x=115 into z-score.
z = (x - mu) / sd
z = (115 - 100) / 6 = 2.5
Now we have to find P(Z > 2.5)
EXCEL syntax to fnd this probability is,
=1 - NORMSDIST(z)
where z is test statistic value.
P(Z > 2.5) = 0.0062
The percentage of demand values would be expected to exceed 115 cans is 0.0062 * 100 = 0.62%.
6. Based upon the population data, what is the demand value that you would expect to be exceeded only 25 percent of the time?
Now we have to find X when percentage is 25% or probability is 25/100 = 0.25
P(X > x) = 0.25
We have to use here inverse probability function for normal distribution.
EXCEL syntax is,
=NORMSINV(probability)
where, probability is 0.25
This will gives us value -0.6745.
(X- mu) / sd = -0.6745
To find value of X plug values of mu and sd,
(X - 100) / 6 = -0.6745
X - 100 = -4.0469
X = 100 - 4.0469
X = 95.9531
The demand value that you would expect to be exceeded only 25 percent of the time is 95.9531.
7. Based upon the population data, what is the demand that you would expect to exceed 90 percent of the time?
Now we have to find X when percentage is 90% or probability is 90/100 = 0.9
P(X > x) = 0.9
We have to use here inverse probability function for normal distribution.
EXCEL syntax is,
=NORMSINV(probability)
where, probability is 0.9
This will gives us value 1.2816.
(X- mu) / sd = 1.2816
To find value of X plug values of mu and sd,
(X - 100) / 6 = 1.2816
X - 100 = 7.6893
X = 100 + 7.6893
X = 107.6893
The demand that you would expect to exceed 90 percent of the time is 107.6893.
Based upon the population data, what is the likelihood of weekly demand being between 85 and 115 cans?
That is we have to find P(85 < X < 115)
z-score for x = 85 and x = 115 is,
z = (85 - 100) / 6 = -2.5
z = (115 - 100) / 6 = 2.5
P(-2.5 < Z < 2.5) = P(Z < 2.5) - P(Z < -2.5) = 0.9938 - 0.0062 = 0.9876
(By using EXCEL syntax is "=NORMSDIST(z) )
The likelihood of weekly demand being between 85 and 115 cans is 0.9876.
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