Question

.) Suppose college students produce 650 pounds of solid waste each year, on average. Assume that the distribution of waste per college student is normal with a mean of 650 pounds and a standard deviation of 20 pounds. What is the probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste

Answer 1

0.073 = 7.3% probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 650 pounds and a standard deviation of 20 pounds.

This means that
`\mu = 650, \sigma = 20`

What is the probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste?

Less than 620:

pvalue of Z when X = 620. So

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

`Z = (620 - 650)/(20)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

More than 700:

1 subtracted by the pvalue of Z when X = 700. So

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

`Z = (700 - 650)/(20)`

`Z = 2.5`

`Z = 2.5` has a pvalue of 0.9938

1 - 0.9938 = 0.0062

Total:

0.0668 + 0.0062 = 0.073

0.073 = 7.3% probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste