A CPA knows from past history that the average accounts receivable for a company is $521.72 with a standard deviation of $584.64. If the auditor takes a simple random sample of 100 accounts, what is the probability that the mean of the sample is within $120 of the population mean

To solve this problem, we need to calculate the probability that the sample mean is within $120 of the population mean. Here's how we can do this step-by-step:

### 1. **Determine the relevant parameters:**

- **Population mean** ($\mu$) = $521.72
- **Population standard deviation** ($\sigma$) = $584.64
- **Sample size** ($n$) = 100
- **Range of interest:** We want to find the probability that the sample mean ($\bar{X}$) is within $120 of the population mean, so the range is:
  \[
  \mu - 120 \leq \bar{X} \leq \mu + 120
  \]

### 2. **Calculate the standard error of the mean (SE):**

The standard error of the mean is given by:
\[
\text{SE} = \frac{\sigma}{\sqrt{n}}
\]
Substituting the values:
\[
\text{SE} = \frac{584.64}{\sqrt{100}} = \frac{584.64}{10} = 58.464
\]

### 3. **Convert the problem to a standard normal distribution:**

We want to find the probability that $\bar{X}$ is within $120 of the population mean. To do this, we calculate the z-scores for the lower and upper bounds of this range.

- **Lower bound z-score:**
\[
z = \frac{(\mu - 120) - \mu}{\text{SE}} = \frac{-120}{58.464} \approx -2.052
\]
  
- **Upper bound z-score:**
\[
z = \frac{(\mu + 120) - \mu}{\text{SE}} = \frac{120}{58.464} \approx 2.052
\]

### 4. **Find the probability using the standard normal distribution:**

We need to find the probability that the z-score is between -2.052 and 2.052. This corresponds to the area under the standard normal curve between these two z-scores.

Using the standard normal distribution table (or a calculator), the probabilities corresponding to these z-scores are:

- $P(Z \leq 2.052) \approx 0.9799$
- $P(Z \leq -2.052) \approx 0.0201$

The probability that the sample mean is within $120 of the population mean is the difference between these probabilities:

\[
P(-2.052 \leq Z \leq 2.052) = 0.9799 - 0.0201 = 0.9598
\]

### **Conclusion:**
The probability that the mean of the sample is within $120 of the population mean is approximately **0.9598** or **95.98%**.

Would you like more details or have any questions?

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Here are some related questions to consider:

1. How would the probability change if the sample size increased to 400?
2. What is the significance of the z-score in the context of the normal distribution?
3. Why is the standard error of the mean smaller than the population standard deviation?
4. How would you calculate the z-score if the range was within $200 of the population mean?
5. What assumptions are necessary for using the normal distribution in this context?
6. How does the Central Limit Theorem apply to this problem?
7. How could you interpret the 95.98% probability in the context of auditing?
8. What would happen if the population distribution was heavily skewed?

**Tip:** Always ensure that the sample size is large enough when using the Central Limit Theorem to approximate the distribution of the sample mean as normal, especially when dealing with non-normal populations.

Calculate the probability that the mean of a sample of 100 accounts is within $120 of the population mean. Understand how to use the normal distribution, standard error of the mean, and z-scores in this statistical problem.

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