Question

An article suggests that substrate concentration (mg/cm3) of influent to a reactor is normally distributed with μ = 0.60 and σ = 0.09. (Round your answers to four decimal places.) (a) What is the probability that the concentration exceeds 0.70? (b) What is the probability that the concentration is at most 0.40? (c) How would you characterize the largest 5% of all concentration values? The largest 5% of all concentration values are above mg/cm3.

Answer 1

a) 13.35% probability that the concentration exceeds 0.70

b) 1.32% probability that the concentration is at most 0.40

c) The largest 5% of all concentration values are above 0.748 mg/cm3.

Explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

`\mu = 0.6, \sigma = 0.09`

(a) What is the probability that the concentration exceeds 0.70?

This is 1 subtracted by the pvalue of Z when X = 0.7. So

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

`Z = (0.7 - 0.6)/(0.09)`

`Z = 1.11`

`Z = 1.11` has a pvalue of 0.8665

1 - 0.8665 = 0.1335

13.35% probability that the concentration exceeds 0.70

(b) What is the probability that the concentration is at most 0.40?

pvalue of Z when X = 0.4. So

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

`Z = (0.4 - 0.6)/(0.09)`

`Z = -2.22`

`Z = -2.22` has a pvalue of 0.0132

1.32% probability that the concentration is at most 0.40

(c) How would you characterize the largest 5% of all concentration values?

They are above the 100-5 = 95th percentile, which is the value of X when Z has a pvalue of 0.95, so X when Z = 1.645.

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

`1.645 = (X - 0.6)/(0.09)`

`X - 0.6 = 1.645*0.09`

`X = 0.748`

The largest 5% of all concentration values are above 0.748 mg/cm3.