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!
