the monthly utility bills in a city are normally distributed, with a mean of $100 and a standard deviation of $12. Find the probability that a randomly selected bill is between $90 to $120

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

answer: approximately 206 bones

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