Question

The pucks used by the National Hockey League for ice hockey must weigh between and ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of ounces and a standard deviation of ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places.

Answer 1

`P(5.5<X<6)=P((5.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P((5.5-5.86)/(0.13)<Z<(6-5.86)/(0.13))=P(-2.769<z<1.077)`

And we can find this probability using the normal standard distribution or excel and we got:

`P(-2.769<z<1.077)=P(z<1.077)-P(z<-2.769)= 0.8593-0.0028 = 0.8565 \approx 0.86`

Explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

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(5.86,0.13)`

Where
`\mu=5.86` and
`\sigma=0.13`

We are interested on this probability

`P(5.5<X<6)`

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(5.5<X<6)=P((5.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P((5.5-5.86)/(0.13)<Z<(6-5.86)/(0.13))=P(-2.769<z<1.077)`

And we can find this probability using the normal standard distribution or excel and we got:

`P(-2.769<z<1.077)=P(z<1.077)-P(z<-2.769)= 0.8593-0.0028 = 0.8565 \approx 0.86`