An observation from a normally distributed population is considered "unusual" if it is more than 2 standard deviations away from the mean. There are several contaminants that can harm a city's water supply. Nitrate concentrations above 10 ppm (parts per million) are considered a health risk for infants less than six month of age. The City of Rexburg reports that the nitrate concentration in the city's drinking water supply is between 1.59 and 2.52 ppm (parts per million,) and values outside of this range are unusual. We will assume 1.59 ppm is the value of mu - 2 sigma and mu + 2 sigma is equal to 2.52 ppm. It is reasonable to assume the measured nitrate concentration is normally distributed.
(Source: City of Rexburg)

Use this information to answer questions 20 and 22.
20. Estimate the mean of the measured nitrate concentration in Rexburg's drinking water. (Round your answer to three decimal places)
21. Estimate the standard deviation of the measured nitrate concentration in Rexburg's drinking water. (Round your answer to three decimal places)
22. Between what two measured nitrate concentrations do approximately 68% of the data values lie?
A.2.055 and 2.520 ppm
B.1.357 and 2.753 ppm
C.1.823 and 2.288 ppm
D.1.590 and 2.520 ppm

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cchilabert cchilabert · Answer 1 · 2020-03-04T04:48:57+01:00

Answer:

Step-by-step explanation:

Hello!

X: nitrate concentration in the city's drinking water supply.

X~N(μ;δ²)

A concentration above 10 ppm is a health risk for infants less than 6 months old.

The reported nitrate concentration is between 1.59 and 2.52 ppm, values outside this range are considered unusual.

If any value of a normal distribution that is μ ± 2δ is considered unusual, it is determined that:

μ - 2δ= 1.59

μ + 2δ= 2.52

20) and 21)

I'll clear the values of the mean and the standard deviation using the given information:

a) μ - 2δ= 1.59 ⇒ μ = 1.59 + 2δ

b) μ + 2δ= 2.52 ⇒ Replace the value of Mu from "a" in "b" and clear the standard deviation:

(1.59 + 2δ) + 2δ= 2.52

1.59 + 4δ= 2.52

4δ= 2.52 - 1.59

δ= 0.93/4

δ= 0.2325

Then the value of the mean is μ = 1.59 + 2(0.2325)

μ = 1.59 + 0.465

μ = 2.055

22)

Acording to the empirical rule, 68% of a normal distribution is between μ±δ

Then you can expect 68% of the distribution between:

μ-δ= 2.055-0.2325= 1.8225

μ+δ= 2.055+0.2325= 2.2875

Correct option: C. 1.823 and 2.288 ppm

I hope it helps!