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,

Let us now write the formula for Z-score,

`z=(x-\mu)/(\sigma)`

Let us solve for z-score,

`\begin{gathered} z=(3500-3350)/(122) \\ =(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%