Question

The alkalinity level of water specimens collected from the Han River in Seoul, Korea, has a mean of 50 milligrams per liter and a standard deviation of 3.2 milligrams per liter (Environmental Science & Engineering, Sept., 1, 2000). Assume that the distribution of alkalinity levels is approximately normal, and find the probability that a water specimen collected from the river has an alkalinity levela. exceeding 45 milligrams per liter.b. below 55 milligrams per liter.c. between 51 and 52 milligrams per liter.

Answer 1

Answer:

a) 0.9406

b) 0.9406

c) 0.1140

Step-by-step explanation:

Given, Mean, μ = 50 mg/L

Standard deviation, σ = 3.2 mg/L

a) The standardized score for 45mg/L is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (45 - 50)/3.2 = - 1.56

To determine the probability of a water specimen having alkalinity greater than 45mg/L, P(x > 45) = P(z ≥ (-1.56))

We'll use data from the normal probability table for these probabilities

P(x > 45) = P(z ≥ (-1.56)) = P(z ≤ 1.56) = P(0 ≤ z ≤ 1.56) + P(z ≤ 0) = 0.4406 + 0.5 = 0.9406

b) The standardized score for 55mg/L is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (55 - 50)/3.2 = 1.56

To determine the probability of a water specimen having alkalinity lesser than 55mg/L, P(x < 55) = P(z ≤ (1.56))

We'll use data from the normal probability table for these probabilities

P(x < 55) = P(z ≤ 1.56) = P(0 ≤ z ≤ 1.56) + P(z ≤ 0) = 0.4406 + 0.5 = 0.9406

c) The standardized score for 51mg/L and 52mg/L each, is their values each minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (51 - 50)/3.2 = 0.31

z = (x - μ)/σ = (52 - 50)/3.2 = 0.63

To determine the probability of a water specimen having alkalinity between 51mg/L and 52mg/L, P(51 ≤ x ≤ 52) = P(0.31 ≤ z ≤ 0.63)

We'll use data from the normal probability table for these probabilities

P(51 ≤ x ≤ 52) = P(0.31 ≤ z ≤ 0.63) = P(0 ≤ z ≤ 0.63) - P(0 ≤ z ≤ 0.31) = 0.2357 - 0.1217 = 0.114

Hope this helps!