Question

A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above 10.983 ounces.

Answer 1

`P(X>10.983)=P((X-\mu)/(\sigma)>(10.983-\mu)/(\sigma))=P(Z>(10.983-10.5)/(0.3))=P(z>1.61)`

And we can find this probability using the complement rule and with excel or the normal standard table:

`P(z>1.61)=1-P(z<1.61)=1-0.946=0.054`

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(10.5,0.3)`

Where
`\mu=10.5` and
`\sigma=0.3`

We are interested on this probability

`P(X>10.983)`

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

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

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

`P(X>10.983)=P((X-\mu)/(\sigma)>(10.983-\mu)/(\sigma))=P(Z>(10.983-10.5)/(0.3))=P(z>1.61)`

And we can find this probability using the complement rule and with excel or the normal standard table:

`P(z>1.61)=1-P(z<1.61)=1-0.946=0.054`