Question

The amount of saturated fat in a daily serving of a particular brand of breakfast cereal is normally distributed with mean 25 g and standard deviation 4 g.

a. Find the sampling distribution of the daily average saturated fat intake over a 30-day period (one month). Include the mean and standard deviation in your answer, as well as the name of the distribution.

b. What is the probability that the average daily saturated fat intake for the month was more than 27 g?

Answer 1

Answer:

a) $\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And replacing we got:

$\bar X \sim N(\mu=25, \frac{4}{\sqrt{30}}= 0.730)$

b) $z =\frac{27-25}{\frac{4}{\sqrt{30}}}= 2.739$

And using the normal standard distribution table and the complement rule we got:

$P(z>2.739) =1- P(z<2.739) = 1-0.997= 0.003$

Step-by-step explanation:

From the info given if we define the random variable X as "amount of saturated fat in a daily serving of a particular brand of breakfast cereal " we know that the distribution of X is given by:

$X \sim N(\mu =25, \sigma =4)$

Part a

For this case the sample size would be n =30 and then the distribution for the sample mean would be given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And replacing we got:

$\bar X \sim N(\mu=25, \frac{4}{\sqrt{30}}= 0.730)$

Part b

We want to find this probability:

$P(\bar X >27)$

And we can use the z score formula given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z =\frac{27-25}{\frac{4}{\sqrt{30}}}= 2.739$

And using the normal standard distribution table and the complement rule we got:

$P(z>2.739) =1- P(z<2.739) = 1-0.997= 0.003$