Question

Fluoride concentration in city water is heavily regulated to ensure there are no adverse long-term health effects. Suppose that a city decides to regulate its fluoride concentration at a mean level 1.0 mg/L, and the concentration of fluoride X is normally distributed with a standard deviation of 0.3 mg/L.

(a) [2 pts] Express the parameters of the distribution of X in proper notation.

(b) [2 pts] What is the probability of observing a fluoride concentration less than 0.7 mg/L?

(c) [4 pts] If ten randomly selected water samples are taken from the city’s water supply, what distribution would the sample mean fluoride concentration follow? In addition to identifying the distribution, specify any parameters for the distribution, check any conditions necessary, and explain why x ̄ follows this distribution.

(d) [2 pts] What is the probability of observing a sample mean fluoride concentration less than 0.7 mg/L in a sample of ten randomly selected water samples?

(e) [3 pts] If a sample of ten water samples results in a sample mean fluoride concentration of 1.2 mg/L, should officials be concerned that there is more fluoride in the water than they want? Explain why or why not.

Answer 1

Here mean level of flouride concentration = = 1.0 mg/L

standard deviation of flouride concentration = = 0.3 mg/L

(a) Here the parameter are given above and if x is the fluride concentration so

x ~ N(1.0 mg/L, 0.3 mg/L)

(b) Pr(x < 0.7 mg/L) = NORMDIST(x < 0.7 mg/L ; 1.0 mg/L ; 0.3 mg/L)

Z = (0.7 - 1)/0.3 = -1

Pr(x < 0.7 mg/L) = NORMDIST(x < 0.7 mg/L ; 1.0 mg/L ; 0.3 mg/L) = Pr(Z < -1) = 0.1587

(c) Here if ten randomly sampled selected water samples are taken from the city’s water supply, the distribution of sample mean fluoride concentration will follow normal distribution with mean = 1.0 mg/L and stanard error = 0.3/sqrt(10) = 0.095 mg/L

Here parameters are

Mean fluoride concentration of sample of 10 = 1.0 mg/L

standard deviation of fluoride concentration of sample of 10 = 0.095 mg/L

~ N(1.0 mg/L ; 0.095 mg/L)

(d) Pr( < 0.7 mg/L) = NORMDIST( < 0.7 mg/L ; 1.0mg/L ; 0.095 mg/L)

Z = (0.7 - 1.0)/0.095 = -3.1623

Pr( < 0.7 mg/L) = NORMDIST( < 0.7 mg/L ; 1.0mg/L ; 0.095 mg/L) = Pr(Z < -3.1623) = 0.0008

(e) here we have to first find the probability of geting mean fluoride concentration of 1.2 mg/L for the sample of 10.

Pr( > 1.2 mg/L) = 1- Pr( < 1.2 mg/L)

= 1 - NORMDIST(< 1.2 mg/L ; 1.0 mg/L ; 0.095 mg/L)

Z = (1.2 - 1.0)/0.095 = 2.1082

Pr( > 1.2 mg/L) = 1 - Pr(Z < 2.1082) = 1- 0.9825 = 0.0175

so, as this probability is less than 0.05, so the officials should be concerned that there is more flouride in the water than they want.