Weekly demand for eggs stocked by Waumini grocers is normally distributed. The mean is 500 trays and standard deviation is 100 trays. How many trays should be available for a week if waumini wants to ensure that the probability of running out of stock does not exceed 4%?
Oh, I see Waumini grocers wants to play it safe and avoid running out of eggs. Don't worry, I'm here to help with some statistical clownery!
To find out how many trays should be available to ensure a probability of running out of stock not exceeding 4%, we need to look at the z-score associated with this probability.
The z-score represents how many standard deviations an observation is from the mean. In this case, we want to find the z-score that corresponds to a probability of 0.04 (4%).
Now, using a z-table or a statistical calculator, we find that the z-score value for a probability of 0.04 is approximately -1.75.
To calculate the number of trays needed, we use the formula:
X = mean + (z-score * standard deviation)
Plugging in the values, we get:
X = 500 + (-1.75 * 100)
X = 500 - 175
X = 325
So, Waumini grocers should make sure they have at least 325 trays of eggs available for the week to ensure a probability of running out of stock does not exceed 4%.
Remember, though, these calculations assume that the weekly demand is normally distributed and there are no other factors influencing the demand for eggs.
To determine how many trays should be available for a week to ensure a probability of running out of stock does not exceed 4%, we can use the z-score formula.
First, let's find the z-score corresponding to a probability of 4%. We can use a standard normal distribution table or a statistical calculator to retrieve this value.
The z-score corresponding to a probability of 4% is approximately -1.75.
Next, we can use the z-score formula to find the corresponding tray value:
Z = (X - μ) / σ
Where:
Z = z-score
X = desired tray value
μ = mean (500 trays)
σ = standard deviation (100 trays)
Plugging in the values, we have:
-1.75 = (X - 500) / 100
Now, we can solve for X:
-1.75 * 100 = X - 500
-175 = X - 500
X = -175 + 500
X = 325
Therefore, to ensure that the probability of running out of stock does not exceed 4%, Waumini should have at least 325 trays available for a week.
To calculate the number of trays that should be available for a week, we need to find the value of the demand that corresponds to the 4th percentile (since we want to ensure that the probability of running out of stock does not exceed 4%).
First, we'll convert the problem to a standard normal distribution by applying the z-score formula:
z = (x - μ) / σ
Where:
x = the demand we want to find
μ = mean
σ = standard deviation
Rearranging the formula to solve for x:
x = z * σ + μ
To find the z-score corresponding to the 4th percentile, we need to find the z-score with an area of 0.04 to the left.
Using a standard normal distribution table or a statistical calculator, we can find that the z-score corresponding to a 4% area to the left is approximately -1.75.
Now, we can calculate the number of trays that should be available:
x = -1.75 * 100 + 500
x = -175 + 500
x = 325
Therefore, Waumini grocers should have at least 325 trays available for a week to ensure that the probability of running out of stock does not exceed 4%.