Question

Using the information below, answer the next 3 questions: 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 about ten. Suppose that 16 individuals are randomly chosen. (a) For the group of 16, find the probability that the average percent of fat calories consumed is more than thirty-five (Round to 3 decimal places)

Answer 1

Solution:

Given that , X follows normal distribution with

$\mu$ = 16

$\sigma$ = 10

A sample of size n = 16 is taken from this population.

Let $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1573406737&ymreqid=50870fb5-0ab3-84d7-1ca9-bb001d01e500&sig=mV2YjzkRFrqEh3lBQGaCHg--~C$ be the mean of sample.

The sampling distribution of the $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1573406737&ymreqid=50870fb5-0ab3-84d7-1ca9-bb001d01e500&sig=mV2YjzkRFrqEh3lBQGaCHg--~C$ is approximately normal with

Mean($\mu_{\bar x}$) = $\mu$ = 16

SD($\sigma_{\bar x}$) =    $\frac{\sigma}{\sqrt{n}}$ = 10/$\sqrt{}$​16 = 2.5

Find P($\bar x$ > 35)

= P[($\bar x$ - $\mu_{\bar x}$ )/$\sigma_{\bar x}$ >  (35 - $\mu_{\bar x}$ )/$\sigma_{\bar x}$]

= P[Z > (35 - 36)/2.5]

= P[Z > -0.40]

= 1 - P[Z < -0.40]

= 1 - 0.3446 ( use z table)

= 0.655

Answer :  0.655