The mean life span of a brand name tire is 50,000 miles. Assume that the life spans of the tires are normally distributed, and the population standard deviation is 800 miles.
a. If you select one tire, what is the probability that its life span is less than 48,500 miles?
b. If you select 100 tires, what is the probability that their mean life span is more than 50,200 miles?

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".

Let X the random variable that represent the mean life span of a brand name tire, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu=50000,\sigma=800)[/tex]

Part a

We want this probability:

[tex]P(X<48500)[/tex]

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(X <48500)=P(Z<\frac{48500-50000}{800})=P(Z<-1.875)=0.0304[/tex]

Part b

Let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

On this case [tex]\bar X \sim N(50000,\frac{800}{\sqrt{100}})[/tex]

We want this probability:

[tex]P(\bar X>50200)=1-P(\bar X<50200)[/tex]

The best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

[tex]P(\bar X >50200)=1-P(Z<\frac{50200-50000}{\frac{800}{\sqrt{100}}})=1-P(Z<2.5)[/tex]

[tex]P(\bar X>50200)=1-0.994=0.0062[/tex]