Question

The trucking company has a policy that any truck with a mileage more than 2.5 standard deviations above the mean should be removed from the road and inspected. What is the probability that a randomly selected truck from the fleet will have to be inspected? ROUND YOUR ANSWER TO 4 DECIMAL PLACES

Answer 1

For this case we want this probability:

`P(X > \mu +2.5 \sigma)`

And we can use the z score formula given by:

`z = ( X -\mu)/(\sigma)`

And replacing we got:

`z = (\mu +2.5 \sigma -\mu)/(\sigma)= 2.5`

We want this probability:

`P(Z>2.5) = 1-P(Z<2.5)`

And using the normal standard table or excel we got:

`P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062`

Explanation:

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 mileage of a population, and for this case we know the distribution for X is given by:

`X \sim N(\mu,\sigma)`

For this case we want this probability:

`P(X > \mu +2.5 \sigma)`

And we can use the z score formula given by:

`z = ( X -\mu)/(\sigma)`

And replacing we got:

`z = (\mu +2.5 \sigma -\mu)/(\sigma)= 2.5`

We want this probability:

`P(Z>2.5) = 1-P(Z<2.5)`

And using the normal standard table or excel we got:

`P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062`

Answer 2

0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected

Explanation:

Z-score

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

What is the probability that a randomly selected truck from the fleet will have to be inspected?

Values of Z above 2.5. So this probability is 1 subtracted by the pvalue of Z = 2.5.

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected