Question

A production line is filling cans of nuts. The label says that the net weight of the can is 16 oz. If the net weight of a can of nuts follows a normal distribution with mean 16.4 oz. and standard deviation of 0.3 oz, what percentage of the cans of nuts will weigh less than the label claims?

5.8%

12.5%

9.2%

2.6%

Answer 1

`P(X<16)=P(\\frac{X-\mu}{\sigma}<(16-\mu)/(\sigma))=P(Z<(16-16.4)/(0.3))=P(Z<-1.33)`

And we can find this probability on this way using the z table or excel:

`P(Z<-1.33)=0.0917`

And that represent approximately 9.2% of the data.

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 weigths of the cans of a population, and for this case we know the distribution for X is given by:

`X \sim N(16.4,0.3)`

Where
`\mu=16.4` and
`\sigma=0.3`

We are interested on this probability

`P(X<16)`

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<16)=P((X-\mu)/(\sigma)<(16-\mu)/(\sigma))=P(Z<(16-16.4)/(0.3))=P(Z<-1.33)`

And we can find this probability on this way using the z table or excel:

`P(Z<-1.33)=0.0917`

And that represent approximately 9.2% of the data.