Question

A regulation hockey puck must weigh between 5.5 and 6 ounces. The weights X of pucks made by a particular process are normally distributed with mean 5.75 ounces and standard deviation 0.11 ounce. Find the probability that a puck made by this process will meet the weight standard.

Answer 1

`P(5.5<X<6)=P((5.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P((5.5-5.75)/(0.11)<Z<(6-5.75)/(0.11))=P(-2.27<z<2.27)`

And we can find this probability with this difference:

`P(-2.27<z<2.27)=P(z<2.27)-P(z<-2.27)=0.988 -0.0116=0.9764`

Explanation:

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(5.75,0.11)`

Where
`\mu=5.75` and
`\sigma=0.11`

We are interested on this probability

`P(5.5<X<6)`

And we can use the z score formula 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((5.5-5.75)/(0.11)<Z<(6-5.75)/(0.11))=P(-2.27<z<2.27)`

And we can find this probability with this difference:

`P(-2.27<z<2.27)=P(z<2.27)-P(z<-2.27)=0.988 -0.0116=0.9764`