Question

The daily demand for meat pies at a specific vendor is normally distributed with a mean of 300 and a standard deviation of 50. How many meat pies should the vendor make to ensure that he runs short on no more than 20% of days?

a. 260
b. 290
c. 320
d. 350

Answer 1

To ensure that the vendor runs short on no more than 20% of days, they should make at least 259 meat pies. The correct answer is option a. 260.

Step-by-step explanation:

In order to find how many meat pies the vendor should make to ensure that he runs short on no more than 20% of days, we need to find the z-score corresponding to the 20th percentile. First, we calculate the z-score using the formula: z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. In this case, x = μ - z(σ), since we want to find the number of pies the vendor should make lower than the 20th percentile. Let's solve for z:

z = (x - μ) / σ = (x - 300) / 50

We can then use a z-table or a calculator to find the z-score corresponding to the 20th percentile. The closest z-score is -0.84. Substituting the known values into the equation, we get:

-0.84 = (x - 300) / 50

Solving for x, we find:

x = -0.84 * 50 + 300 = 258

So, the vendor should make at least 259 meat pies to ensure that he runs short on no more than 20% of days. Therefore, the correct answer is option a. 260.