the monthly utility bills in a city are normally distributed, with a mean of $100 and a standard deviation of $12. Find the prob
1 answer:
Answer:
0.7493 or 74.93%
Step-by-step explanation:
We are given;
Population mean; μ = 100
Population standard deviation; σ = 12
Sample mean 1; x1¯ = 90
Sample mean 2; x2¯ = 120
Z-score formula of the data given is;
z = (x¯ - μ)/σ
z1 = (90 - 100)/12
z1 = -0.83
z2 = (120 - 100)/12
z2 = 1.67
Since we wan to Find the probability that a randomly selected bill is between $90 to $120.
Thus;
P(90 < x¯ < 120) = P(−0.83 < Z < 1.67)
Probability will be gotten from online probability with 2 z-scores calculator to get;
P = 0.7493
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