Respuesta :
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!
