Question

The weights of ice cream cartons are normally distributed with a mean weight of 1313 ounces and a standard deviation of 0.50.5 ounce. ​(a) What is the probability that a randomly selected carton has a weight greater than 13.1713.17 ​ounces? ​(b) A sample of 3636 cartons is randomly selected. What is the probability that their mean weight is greater than 13.1713.17 ​ounces?

Answer 1

Answer:

a) We are interested on this probability

$P(X>13.17)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367$

b) $P(\bar X >13.17)$

The new z score is given by:

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

And replacing we got:

$P(\bar X >13.17)=P(Z> \frac{13.17-13}{\frac{0.5}{\sqrt{36}}})= P(Z>2.04) =0.0207$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(13,0.5)$  

Where $\mu=13$ and $\sigma=0.5$

We are interested on this probability

$P(X>13.17)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367$

Part b

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

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

And we want this probability:

$P(\bar X >13.17)$

The new z score is given by:

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

And replacing we got:

$P(\bar X >13.17)=P(Z> \frac{13.17-13}{\frac{0.5}{\sqrt{36}}})= P(Z>2.04) =0.0207$