store manager is doing a study on how to staff his franchise. He knows that the average number of pizzas ordered is 12 per hour with a standard deviation of 1.4 pizzas. Staffed with 3 employees, he can assemble up to 15 pizzas per hour. If his store is open for 10 hours each day, what is the probability that he will receive more orders than he can assemble in a given hour?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n(hours)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To study the probability that the store manager will receive more orders than he can assemble in a given hour, we need to compare the average number of pizzas ordered per hour with the number of pizzas the store can assemble in that hour.
The average number of pizzas ordered per hour is 12, with a standard deviation of 1.4 pizzas. This gives us information about the distribution of pizza orders.
The store can assemble up to 15 pizzas per hour, given that it is staffed with 3 employees.
To calculate the probability of the manager receiving more orders than he can assemble in a given hour, we need to find the area under the distribution curve that represents the probability of ordering more than 15 pizzas in an hour.
1. Calculate the z-score: The z-score measures the number of standard deviations away from the mean a particular value is. In this case, we want to find the z-score for 15 pizzas. The formula for calculating the z-score is:
z = (x - μ) / σ
Where x is the value (15 pizzas), μ is the mean (12 pizzas), and σ is the standard deviation (1.4 pizzas).
z = (15 - 12) / 1.4 = 2.14
2. Find the probability: Once we have the z-score, we can use a standard normal distribution table or a statistical calculator to find the probability associated with that z-score. The probability is the area under the curve to the right of the z-score.
The probability can be denoted as P(z > 2.14), which means the probability of ordering more than 15 pizzas in an hour.
By looking up the z-score of 2.14 in a standard normal distribution table or using a statistical calculator, we find that the probability is approximately 0.016.
Therefore, the probability that the store manager will receive more orders than he can assemble in a given hour is approximately 0.016, or 1.6%.