Question

The percent of fat calories that a person consumes each day is normally distributed with a mean of 37 and a standard deviation of 10. Suppose that 16 individuals are randomly chosen. Let X = average percent of fat calories. (a) For the group of 16, find the probability that the average percent of fat calories consumed is more than forty. (Round your answer to four decimal places.) 0.382 b) Find the first quartile for the average percent of fat calories. (Round your answer to two decimal places.) 30.25 percent of fat calories

Answer 1

Answer:

a) 38.21% probability that the average percent of fat calories consumed is more than forty.

b) The first quartile is 30.25% of fat 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:

The percent of fat calories that a person consumes each day is normally distributed with a mean of 37 and a standard deviation of 10, so $\mu = 37, \sigma = 10$.

(a) For the group of 16, find the probability that the average percent of fat calories consumed is more than forty.

This is the 1 subtracted by the pvalue of Z when $X = 40$.

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

$Z = \frac{40 - 37}{10}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179.

So, there is a 1-0.6179 = 0.3821 = 38.21% probability that the average percent of fat calories consumed is more than forty.

b) Find the first quartile for the average percent of fat calories. (Round your answer to two decimal places.) 30.25 percent of fat calories

What is the value of Z that has a pvalue of 0.25? That is between $Z = -0.67$ and $Z = -0.68$, so we use $Z= -0.675$.

So, we have to find the value of X when $Z = -0.675$.

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

$-0.675 = \frac{X - 37}{10}$

$X - 37 = -6.75$

$X = 30.25$

The first quartile is 30.25% of fat calories.