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Let's solve each part of the question using properties of the normal distribution.

Given:
- Mean ($\mu$) = $40,000
- Standard Deviation ($\sigma$) = $1,500

### (a) About 68% of the incomes lie between what two amounts?
In a normal distribution, about 68% of the data lies within one standard deviation of the mean. This means:

\[
\text{Lower bound} = \mu - \sigma = 40,000 - 1,500 = 38,500
\]
\[
\text{Upper bound} = \mu + \sigma = 40,000 + 1,500 = 41,500
\]

So, 68% of the incomes lie between **$38,500 and $41,500**.

### (b) About 95% of the incomes lie between what two amounts?
Approximately 95% of the data in a normal distribution falls within two standard deviations of the mean. This means:

\[
\text{Lower bound} = \mu - 2\sigma = 40,000 - 2(1,500) = 37,000
\]
\[
\text{Upper bound} = \mu + 2\sigma = 40,000 + 2(1,500) = 43,000
\]

So, 95% of the incomes lie between **$37,000 and $43,000**.

### (c) What is the median income?
In a normal distribution, the mean is equal to the median. Therefore, the median income is also **$40,000**.

### (d) Is the distribution of incomes symmetrical?
Yes, the distribution is symmetrical. In a normal distribution, data is symmetrically distributed around the mean, with equal amounts of data on both sides.

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Would you like further details on any of these answers or have any questions?

Here are some additional questions for practice:
1. What percentage of incomes lie between $36,500 and $43,500?
2. What is the probability that a randomly selected income is more than $43,000?
3. If the standard deviation was $2,000 instead of $1,500, what would the range for 95% of incomes be?
4. How would the range for 68% of incomes change if the mean increased to $45,000?
5. What income level corresponds to one standard deviation below the mean?

**Tip:** For normally distributed data, always remember the empirical rule: approximately 68% within 1 standard deviation, 95% within 2, and 99.7% within 3.

The distribution of the annual incomes of a group of middle-management employees at XYZ company approximates a normal distribution with a mean of $40,000 and a standard deviation of $1500.
(a) About 68% of the incomes lie between what two amounts?
(b) About 95% of the incomes lie between what two amounts?
(c) What is the median income?
(d) Is the distribution of incomes symmetrical?

Analyze the normal distribution of annual incomes among XYZ company's middle-management employees, with a mean of $40,000 and standard deviation of $1,500. Learn how to determine the income ranges for 68% and 95% of employees, understand the median income, and assess the symmetry of the distribution. Suitable for high school students studying statistics.

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