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In understanding uniform distribution for a given range, the first step involves calculating the mean, which also represents the average of that distribution. It tells us the central tendency or the midpoint for a uniformly distributed random variable.

For a uniform distribution within a range from a minimum value, \(a\), to a maximum value, \(b\), the mean is determined using the formula:

• \( \mu = \frac{a + b}{2} \)

This straightforward formula encapsulates the concept that, for uniform distribution, every point within the given range is equally likely. In this example, with spending ranging from \(\\(400\) to \(\\)3,800\), the mean expenditure is calculated as:

• \( \mu = \frac{400 + 3800}{2} = 2100 \)

Hence, a family of four is expected to spend, on average, \(\$2,100\) annually on insurance.

Standard deviation is a measure of the spread of data points in a distribution. In uniform distribution, it explains how much the individual points may deviate from the mean within a specified range.

For a uniform distribution, the formula for standard deviation, \(\sigma\), is given by:

• \( \sigma = \frac{b - a}{\sqrt{12}} \)

In the context of our example, where the insurance expenditure ranges from \(\\(400\) to \(\\)3,800\), the standard deviation is calculated as:

• \( \sigma = \frac{3800 - 400}{\sqrt{12}} \approx 980.58 \)

This figure indicates that the spending by each family is distributed quite widely around the mean (\(\$2,100\)), reflecting the variability in expenditure levels.

Probability in the context of uniform distribution allows us to calculate the likelihood of an event occurring within a specific interval. For this distribution, each interval of the same length has an equal chance of occurring.

The probability that a randomly selected family spends less than \(\\(2,000\) is determined by the fraction of the total interval where the spending is below \(\\)2,000\).

• The interval from \(\\(400\) to \(\\)2,000\) has a length of \(1,600\).
• The total interval \([400, 3800]\) has a length of \(3,400\).

Thus the probability is:

• \( P(X < 2000) = \frac{1600}{3400} \approx 0.471 \)

Similarly, the probability of spending more than \(\$3,000\) is calculated by the length of the interval \([3,000, 3,800]\), which is \(800\), thus:

• \( P(X > 3000) = \frac{800}{3400} \approx 0.235 \)

These probabilities provide insights into the spending behaviors of families within the given range.

Insurance expenditure refers to the amount of money a family spends on various types of insurance each year. Understanding the distribution of this expense among families can provide insights into average financial commitments and potential savings opportunities.

Given the uniformly distributed expenditure data, we ascertain the mean and standard deviation to understand the average and variability in spending. Moreover, probabilities help predict behaviors in spending.

• If the mean is \(\\(2,100\), it indicates a central point of expenditure.
• A standard deviation near \(\\)980.58\) shows potential large swings from the mean, providing a snapshot of spending inconsistency.
• Probabilities show the likelihood of a family spending within a certain range.

It's crucial in everyday decision-making for budgeting and can influence financial planning for families wanting to optimize their insurance purchases. By understanding these statistics, families can better position themselves to either maintain or adjust their insurance plans.