Question

In 2013, a study of restaurant meals in Boston, MA found that on average, single plate meals contained 1,327 calories. Per the USDA, men require at most 2,000 to 3,000 calories per day (distributed across the day). Assuming restaurant meal caloric content is normally distributed with a standard deviation of 500. There is an interval in which 90% of all meals are symmetrically distributed about the mean. What is the upper limit of that interval?

Answer 1

Answer:

The upper limit of that is 2149.5 calories.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that

Single plate meals contained 1,327 calories, so $\mu = 1327$

Assuming restaurant meal caloric content is normally distributed with a standard deviation of 500, so $\sigma = 500$

There is an interval in which 90% of all meals are symmetrically distributed about the mean. What is the upper limit of that interval?

This interval is from a pvalue of 0.05(lower limit) to a pvalue of 0.95(upper limit).

So, the first step is to find the value of Z that has a pvalue of 0.95. This is between $Z = 1.64$ and $Z = 1.65$. So i am going to solve for $Z = 1.645$.

Now, we apply the formula to find the value of X.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 1327}{500}$

$X - 1327 = 822.5$

$X = 2149.5$

The upper limit of that is 2149.5 calories.