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

Answer:

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

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.

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 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:

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

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

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

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.