Monthly cell phone bills for residents of a city are approximately normally distributed and have a mean of $92 and standard deviation of $12. Simple random samples of 100 are drawn and the mean is determined for each sample. What is the probability that an individual resident will have a monthly cell phone bill of $100 or more?

To determine the probability that an individual resident will have a monthly cell phone bill of $100 or more, we'll use the properties of the normal distribution.

Given:
- The mean ($\mu$) of the monthly bills = $92
- The standard deviation ($\sigma$) of the monthly bills = $12

We need to find the probability that an individual bill exceeds $100. This involves calculating the area under the normal distribution curve to the right of $100.

First, we convert the amount of $100 to a Z-score using the formula:

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

Where:
- $X$ is the value of interest (here, $100)
- $\mu$ is the mean
- $\sigma$ is the standard deviation

Plugging in the values:

\[
Z = \frac{100 - 92}{12} = \frac{8}{12} = \frac{2}{3} \approx 0.667
\]

Next, we find the probability that a standard normal variable is greater than this Z-score. We look up the Z-score of 0.667 in the standard normal distribution table or use a calculator.

The cumulative probability up to a Z-score of 0.667 is approximately 0.7475. Since we want the probability of exceeding $100, we need the area to the right of this Z-score, which is:

\[
P(X > 100) = 1 - P(Z \leq 0.667)
\]

\[
P(X > 100) = 1 - 0.7475 = 0.2525
\]

Thus, the probability that an individual resident will have a monthly cell phone bill of $100 or more is approximately **0.2525** (or 25.25%).

Would you like further details or have any questions?

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Here are some related questions for further exploration:

1. What is the probability that an individual resident will have a bill less than $80?
2. What is the probability that an individual resident's bill is between $85 and $105?
3. How would the probability change if the standard deviation was reduced to $8?
4. What proportion of bills fall between $90 and $94?
5. What is the 95th percentile of the monthly bills?
6. How would we calculate the probability of the mean bill being over $95 for a sample of 100 residents?
7. What is the standard error of the mean for samples of size 100?
8. How does the central limit theorem apply to this problem?

**Tip:** When working with normally distributed data, always consider using Z-scores to standardize the values for easier calculation of probabilities.

Learn how to calculate the probability that an individual resident will have a monthly cell phone bill of $100 or more using the properties of normal distribution. Given mean $92 and standard deviation $12, find the probability with detailed step-by-step solution.

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