Problem 4 - Basic Statistics for Business and Economics

Question

According to the Insurance Institute of America, a family of four spends between \(\$ 400\) and \(\$ 3,800\) per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts. a. What is the mean amount spent on insurance? b. What is the standard deviation of the amount spent? c. If we select a family at random, what is the probability they spend less than \(\$ 2,000\) per year on insurance per year? d. What is the probability a family spends more than \(\$ 3,000\) per year?

Step-by-Step Solution

Answer

Mean: $2,100; Std Dev: $981.47; P(<$2,000): 47.06%; P(>$3,000): 23.53%.

Step 1: Understanding Uniform Distribution

For this problem, we have a continuous uniform distribution, with given minimum (a) of $400 and maximum (b) of $3,800. This means each value between $400 and $3,800 has an equal probability of occurring.

Step 2: Calculating the Mean

The mean of a uniform distribution is calculated as \[ \text{mean} = \frac{a+b}{2} \] Substituting the given values, \[ \text{mean} = \frac{400 + 3800}{2} = \frac{4200}{2} = 2100 \] So, the mean amount spent on insurance is $2,100.

Step 3: Calculating the Standard Deviation

The standard deviation of a uniform distribution is calculated using the formula \[ \text{standard deviation} = \sqrt{\frac{(b-a)^2}{12}} \] Substituting the given values, \[ \text{standard deviation} = \sqrt{\frac{(3800-400)^2}{12}} = \sqrt{\frac{3400^2}{12}} = \sqrt{\frac{11560000}{12}} \approx \sqrt{963333.33} \approx 981.47 \] Thus, the standard deviation is approximately $981.47.

Step 4: Calculating Probability of Spending Less than $2,000

For a uniform distribution, the probability that a variable X is less than a certain value c is given by \[ P(X < c) = \frac{c-a}{b-a} \] Here, c = \(2,000. Thus, \[ P(X < 2000) = \frac{2000-400}{3800-400} = \frac{1600}{3400} \approx 0.4706 \] Therefore, the probability a family spends less than \)2,000 is approximately 0.4706 or 47.06%.

Step 5: Calculating Probability of Spending More than $3,000

For a uniform distribution, the probability that a variable X is more than a certain value d can be found by \[ P(X > d) = 1 - P(X < d) \] For d = \(3,000, we first calculate \( P(X < 3000) \): \[ P(X < 3000) = \frac{3000-400}{3800-400} = \frac{2600}{3400} \approx 0.7647 \] Thus, \[ P(X > 3000) = 1 - 0.7647 = 0.2353 \] Therefore, the probability a family spends more than \)3,000 is approximately 0.2353 or 23.53%.

Key Concepts

Expected Value

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

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

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.