Exercise
Answer
Solution: D Let [math]X[/math] denote the number of deaths next year among the 1000 employees. Each employee has a 1.4% chance of dying, independently. This means [math]X[/math] follows a binomial distribution.
| Parameter | Value |
|---|---|
| Number of Employees ([math]n[/math]) | 1000 |
| Probability of Death ([math]p[/math]) | 0.014 |
Therefore, [math]X \sim \textrm{Bin}(1000, 0.014)[/math]. Let [math]S[/math] denote the total life insurance payments next year. Each death results in a benefit of $50,000. So, the total payment is [math]S = 50,000X[/math].
The expected number of deaths is given by the formula for a binomial distribution: [math]\operatorname{E}(X) = np[/math].
The variance of the number of deaths is given by the formula for a binomial distribution: [math]\operatorname{Var}(X) = np(1-p)[/math].
Since the number of employees ([math]n = 1000[/math]) is large, we can approximate the binomial distribution of [math]X[/math] (and thus the distribution of [math]S[/math]) using the Central Limit Theorem with a normal distribution. We need to find the fund amount [math]F[/math] such that the probability that the fund will cover next year's death benefits is at least 0.99. This is expressed as [math]P(S \le F) \ge 0.99[/math]. This means [math]F[/math] is the 99th percentile of the distribution of [math]S[/math]. For a standard normal distribution, the Z-score corresponding to the 99th percentile is approximately [math]z_{0.99} = 2.326[/math]. The formula to calculate the fund amount [math]F[/math] using the normal approximation is:
The calculated fund amount is approximately $1,132,099. We need to round this amount to the nearest 50 thousand.
- The nearest multiples of 50,000 are:
- [math]1,100,000 = 22 \times 50,000[/math]
- [math]1,150,000 = 23 \times 50,000[/math]
Since $1,132,099 is closer to $1,150,000 (difference of $17,901) than to $1,100,000 (difference of $32,099), the smallest amount the company must put into the fund, rounded to the nearest 50 thousand, is $1,150,000.
- Understanding the Random Variable: The total claim amount is a multiple of a binomial random variable (number of deaths). Correctly identifying the distribution and its parameters is crucial.
- Binomial Distribution Properties: For a binomial distribution [math]\textrm{Bin}(n, p)[/math], the expected value is [math]np[/math] and the variance is [math]np(1-p)[/math]. These fundamental formulas are often starting points for such problems.
- Scaling Expected Value and Variance: When a random variable is scaled by a constant (e.g., [math]S = cX[/math]), its expected value scales linearly ([math]\operatorname{E}(S) = c\operatorname{E}(X)[/math]), and its variance scales by the square of the constant ([math]\operatorname{Var}(S) = c^2\operatorname{Var}(X)[/math]). Consequently, the standard deviation scales linearly, [math]\operatorname{StdDev}(S) = |c|\operatorname{StdDev}(X)[/math].
- Central Limit Theorem (CLT) Application: For a sufficiently large number of trials [math]n[/math], a binomial distribution can be approximated by a normal distribution. This allows for the use of Z-scores to calculate percentiles, which is common in risk management and actuarial science for determining reserve levels.
- Determining Fund Levels for a Given Probability: To ensure a fund covers liabilities with a specified probability (e.g., 99%), one must calculate the corresponding percentile of the liability distribution. This often involves the formula [math]\text{Fund} = \operatorname{E}(S) + z_{\alpha} \times \operatorname{StdDev}(S)[/math], where [math]z_{\alpha}[/math] is the Z-score for the desired confidence level [math]\alpha[/math].
- Practical Rounding: Financial calculations often require rounding to specific increments (e.g., nearest thousand, nearest 50 thousand) rather than standard mathematical rounding. It's important to understand the business context for rounding rules.