Question

The weight of a product is normally distributed with a mean of four ounces and a variance of .25 squared ounces. What is the probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces

Answer 1

Answer:

$P(X>3.75)=P(\frac{X-\mu}{\sigma}>\frac{3.75-\mu}{\sigma})=P(Z>\frac{3.75-4}{0.5})=P(z>-0.5)$

And we can find this probability using the complement rule and we got:

$P(z>-0.5)=1-P(z<-0.5)=1-0.309=0.691$

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".  

Solution to the problem

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:

$X \sim N(4,\sqrt{0.25}=0.5)$  

Where $\mu=4$ and $\sigma=0.5$

We are interested on this probability

$P(X>3.75)$

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:

$P(X>3.75)=P(\frac{X-\mu}{\sigma}>\frac{3.75-\mu}{\sigma})=P(Z>\frac{3.75-4}{0.5})=P(z>-0.5)$

And we can find this probability using the complement rule and we got:

$P(z>-0.5)=1-P(z<-0.5)=1-0.309=0.691$