Question

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X= percent of fat calories.(a) Find the z-score corresponding to 30 percent of fat calories, rounded to 3 decimal places. (b) Find the probability that the percent of fat calories a person consumes is more than 30. (c) Find the maximum number for the lower quarter of percent of fat calories. Round your answer to 3 decimal places.

Answer 1

SOLUTION

Write out the given parameters

$\begin{gathered} \mu=36 \\ \sigma=10 \\ x=30 \end{gathered}$

a). The Z-score is given by

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{30-36}{10} \\ \end{gathered}$

Then

$z=-\frac{6}{10}=-0.600$

Hence

The Z-score value corresponding to 30 % of fat calories is -0.600

b). The probability that the percent of fat calories a person consumes is more than 30 is given by

$p(x>30)=p(z>-0.6)=0.7257469$

Hence

The probability that the percent of fat calories a person consumes is more than 30 is 0.7257