Question

The amount of pollutants released from a smoke stack in a day is a normal random variable with mean 4.5 ounces and standard deviation 0.54 ounces. We observe the amount of pollutants released over randomly selected 13 days.
a. What is the expected value of average pollutant released over the selected 13 days? i.e. E([(x)]) = ? [Answer to 2 decimal places]

b. What is the standard deviation of average pollutant released over the selected 13 days? i.e. σ([(x)]) = ? [Answer to 4 decimal places]

c. What is the probability that the average release over 13 days is within 0.25 ounces of the mean? i.e. P(μ− 0.25 ≤ [(x)] ≤ μ+ 0.25) = ? [Answer to 4 decimal places]

Answer 1

a)

E(x) = 4.5

b)

S(x) = s/sqrt(n)
= 0.54/sqrt(13)
= 0.1498

c)

Here, μ = 4.5, σ = 0.1498, x1 = 4.25 and x2 = 4.75. We need to compute P(4.25<= X <= 4.75). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (4.25 - 4.5)/0.1498 = -1.67
z2 = (4.75 - 4.5)/0.1498 = 1.67

Therefore, we get
P(4.25 <= X <= 4.75) = P((4.75 - 4.5)/0.1498) <= z <= (4.75 - 4.5)/0.1498)
= P(-1.67 <= z <= 1.67) = P(z <= 1.67) - P(z <= -1.67)
= 0.9525 - 0.0475
= 0.9050