Question

ABC trucking company realized that on an annual basis that distance traveled is normally distrubuted with a mean of 50,000 miles and a standard deviation of 12,000 HOw many miles will be traveled by at least 80% of the trucks

Answer 1

Answer:

39896 miles will be traveled by at least 80% of the trucks

Step-by-step explanation:

Given that :

the mean  $\mu$ = 50000

standard deviation  $\sigma$ = 12000

we are to calculate how many miles  will be traveled by at least 80% of the trucks.

This implies that :

$P(X > x_o) = 0.8$

Likewise;

$P(X < x_o) = 1- P(X > x_o)$

$P(X < x_o) = 1-0.8$

$P(X < x_o) = 0.2$

We all know that

$z = \dfrac{X- \mu}{\sigma}$

$P(\dfrac{X- \mu}{\sigma}< \dfrac{x_o - \mu }{\sigma}) = 0.2$

$P({z < \dfrac{x_o - \mu }{\sigma}) = 0.2$

Using the z table to determine the value for (invNorm (0.2)); we have ;

$\dfrac{x_o - \mu }{\sigma} = invNorm (0.2)$

${x_o - \mu } = {\sigma} \times invNorm (0.2)$

${x_o } = \mu + {\sigma} \times invNorm (0.2)$

From z tables; $invNorm (0.2)= -0.842$

${x_o } = 50000 + 12000 \times(-0.842)$

${x_o } = 50000 -10104$

$\mathbf{{x_o } =39896}$

Thus; 39896 miles will be traveled by at least 80% of the trucks