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 the sample mean cell phone bill is $100 or more?

To find the probability that the sample mean of the monthly cell phone bills for a sample of 100 residents is $100 or more, we use the concept of the sampling distribution of the sample mean. The central limit theorem tells us that the distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large, even if the population distribution is not normal.

Given:
- Population mean ($\mu$) = $92
- Population standard deviation ($\sigma$) = $12
- Sample size ($n$) = 100

First, we need to calculate the standard error of the mean (SEM), which is the standard deviation of the sampling distribution of the sample mean. The SEM is given by:

\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\]

Plugging in the values:

\[
\text{SEM} = \frac{12}{\sqrt{100}} = \frac{12}{10} = 1.2
\]

Next, we calculate the Z-score for a sample mean of $100 using the formula:

\[
Z = \frac{\bar{X} - \mu}{\text{SEM}}
\]

Where:
- $\bar{X}$ is the sample mean of interest ($100)
- $\mu$ is the population mean ($92)
- $\text{SEM}$ is the standard error of the mean (1.2)

Plugging in the values:

\[
Z = \frac{100 - 92}{1.2} = \frac{8}{1.2} \approx 6.67
\]

Now, we find the probability that the Z-score is greater than 6.67. Since this Z-score is extremely high, it corresponds to a very small probability. The cumulative probability for a Z-score of 6.67 is virtually 1, meaning the probability of the sample mean being $100 or more is very close to 0.

Thus, the probability that the sample mean cell phone bill is $100 or more is effectively **0** (or, more precisely, an extremely small value close to zero).

Would you like more details or have any questions?

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Here are some related questions for further exploration:

1. What is the probability that the sample mean is less than $85 for samples of 100 residents?
2. How would the standard error of the mean change if the sample size were increased to 400?
3. What is the probability that the sample mean falls between $90 and $94?
4. How does the central limit theorem help in determining the distribution of the sample mean?
5. What is the probability that the sample mean is exactly equal to $92?
6. If the standard deviation of the population increased, how would that affect the standard error?
7. How does the sampling distribution of the sample mean differ from the population distribution?
8. What would the Z-score be for a sample mean of $95?

**Tip:** Remember, the larger the sample size, the smaller the standard error, which makes the distribution of the sample mean narrower and more precise around the population mean.

Explore the probability that the sample mean of monthly cell phone bills for a sample of 100 residents is $100 or more. Understand the central limit theorem, Z-scores, and the sampling distribution of the sample mean. Calculations based on a population with mean $92 and standard deviation $12.

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