Answer :
Answer:
[tex]P(5.5<X<6)=P(\frac{5.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5.5-5.86}{0.13}<Z<\frac{6-5.86}{0.13})=P(-2.769<z<1.077)[/tex]
And we can find this probability using the normal standard distribution or excel and we got:
[tex]P(-2.769<z<1.077)=P(z<1.077)-P(z<-2.769)= 0.8593-0.0028 = 0.8565 \approx 0.86[/tex]
Step-by-step 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:
[tex]X \sim N(5.86,0.13)[/tex]
Where [tex]\mu=5.86[/tex] and [tex]\sigma=0.13[/tex]
We are interested on this probability
[tex]P(5.5<X<6)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(5.5<X<6)=P(\frac{5.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5.5-5.86}{0.13}<Z<\frac{6-5.86}{0.13})=P(-2.769<z<1.077)[/tex]
And we can find this probability using the normal standard distribution or excel and we got:
[tex]P(-2.769<z<1.077)=P(z<1.077)-P(z<-2.769)= 0.8593-0.0028 = 0.8565 \approx 0.86[/tex]