Questions: The monthly utility bills in a city are normally distributed with a mean of 100 and a standard deviation of 16. Find the probability that a randomly selected utility bill is less than 87. (Round to four decimal places as needed.)

Solution Steps

To find the probability that a randomly selected utility bill is less than $87, we first calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 87 \) (the value we are interested in),
• \( \mu = 100 \) (the mean of the distribution),
• \( \sigma = 16 \) (the standard deviation of the distribution).

Substituting the values, we have:

\[ z = \frac{87 - 100}{16} = \frac{-13}{16} = -0.8125 \]

Thus, the Z-score for the value 87 is \( z = -0.8125 \).

Next, we calculate the probability that a randomly selected utility bill is less than $87. This is given by:

\[ P(X < 87) = \Phi(Z_{end}) - \Phi(Z_{start}) \]

In this case, \( Z_{end} = -0.8125 \) and \( Z_{start} = -\infty \). Therefore, we have:

\[ P(X < 87) = \Phi(-0.8125) - \Phi(-\infty) \]

Since \( \Phi(-\infty) = 0 \), we can simplify this to:

\[ P(X < 87) = \Phi(-0.8125) \]

From the calculations, we find that:

\[ P(X < 87) \approx 0.2083 \]

Final Answer