Question

The life expectancy of a certain tire is a normal distribution with the mean = 40,000 miles, standard deviation is 1,000 miles. Find the probability that a tire will last more than 39,000 miles.

Answer 1

Answer:

$P = 84\%$

Step-by-step explanation:

The average is:

$\mu = 40,000\ miles$

The standard deviation is:

$\sigma = 1,000\ miles$

We want the probability that a tire lasts more than 39,000 miles.

This is:

$P (X>39,000)$

Now we must transform these values to those of a standard normal distribution to facilitate calculation by using the probability tables.

$P (X> 39,000)\\\\P (X-\mu> 39,000-40,000)\\\\P (\frac{X-\mu}{\sigma}> \frac{39,000 -40,000}{1,000})\\\\P (Z> -1)$

This is:

$P(Z> -1) = P(Z<1)$ ---------- (For the symmetry of the standard normal distribution)

When you search for the normal standard table, you get the following value:

$P(Z <1) = 0.8413\\\\P(Z> -1) = 0.8413$