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The expected value in the context of a uniform distribution represents the mean or average value we anticipate a random variable to be. It's essentially a balancing point of our data.

To determine the expected value (or mean) for money spent on insurance by a family, we use the formula:

• \[ \text{mean} = \frac{a+b}{2} \]

where \( a \) is the lower limit and \( b \) is the upper limit of the distribution. In this scenario, \( a = 400 \) and \( b = 3800 \). Utilizing this formula:

• \[ \text{mean} = \frac{400 + 3800}{2} = 2100 \]

This suggests that, on average, a family of four spends $2,100 annually on insurance. This calculation gives us a central value around which we can expect the spending amounts to vary hypothetically. Remember, this is a theoretical prediction based on uniform distribution assumptions.

Standard deviation, in the simplest terms, measures how spread out the numbers are within a dataset. For a uniform distribution, it specifically tells us how much the values deviate from the mean.

To find the standard deviation in uniform distribution, we use:

• \[ \text{standard deviation} = \sqrt{\frac{(b-a)^2}{12}} \]

Where again, \( a \) is 400 and \( b \) is 3800. By substituting these values into the formula, we get:

• \[ \text{standard deviation} = \sqrt{\frac{(3800-400)^2}{12}} \approx 981.47 \]

This indicates that the amount a family spends on insurance can deviate from the mean (expected value) by about $981.47. A higher standard deviation here means that the insurance spending habits between families can vary significantly.

Probability helps us understand how likely an event is to happen. In this context, we're looking to find probabilities associated with how much money families spend on insurance. For a uniform distribution, calculating probabilities involves comparing the range of interest with the total possible range.When calculating the probability of a family spending less than a certain amount, like \(2,000, we use:

• \[ P(X < c) = \frac{c-a}{b-a} \]

where \( c \) is the amount of interest. Substituting \( c = 2000 \):

• \[ P(X < 2000) = \frac{1600}{3400} \approx 0.4706 \]

This tells us that there's a 47.06% chance a family spends less than \)2,000 per year.For the probability of spending more than \(3,000, we find it by seeing how often they spend less than \)3,000 and subtracting from 1:

• \[ P(X > 3000) = 1 - P(X < 3000) = 1 - 0.7647 = 0.2353 \]

This gives a probability of 23.53% that a randomly selected family spends more than $3,000. Probability in uniform distributions hence gives us a straightforward way of predicting outcomes based on given data.