Question

The number of calories per day consumed by adults is normally distributed with a mean of 3350 and with a standard deviation of 122. Which of the following is closest to the percent of adults who eat more than 3500 calories per day?

Answer 1

Let us write out the given data,

$\begin{gathered} \mu=\operatorname{mean}=3350 \\ \sigma=standard\text{ deviation=122} \\ x=3500 \\ z=z-\text{score} \end{gathered}$

Let us now write the formula for Z-score,

$z=\frac{x-\mu}{\sigma}$

Let us solve for z-score,

$\begin{gathered} z=\frac{3500-3350}{122} \\ =\frac{150}{122}=1.2295 \\ z=1.2295 \end{gathered}$

The probablity that the percent of adult who eat more than 3500 will be

[tex]\begin{gathered} Pr(z>1.23)\Rightarrow Pr(0Hence,the percent of adults who eat more than 3500 calories per day is 11%