SOLUTION: What is the paid-up life insurance value of a $50,000 20-year limited-payment life insurance policy that has been owned for 10 years, if the value per $1,000 unit is $428?
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: What is the paid-up life insurance value of a $50,000 20-year limited-payment life insurance policy that has been owned for 10 years, if the value per $1,000 unit is $428?
Log On
Question 1179340: What is the paid-up life insurance value of a $50,000 20-year limited-payment life insurance policy that has been owned for 10 years, if the value per $1,000 unit is $428? Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! You've got the correct approach! Here's a recap of the steps and the final answer:
1. Define the Variables:
Mean (μ) = 287 grams
Standard deviation (σ) = 19 grams
Lower bound (x1) = 251 grams
Upper bound (x2) = 346 grams
2. Calculate the Z-Scores:
We need to convert the given weights to z-scores using the formula:
z = (x - μ) / σ
For x1 = 251 grams:
z1 = (251 - 287) / 19 = -36 / 19 ≈ -1.89
For x2 = 346 grams:
z2 = (346 - 287) / 19 = 59 / 19 ≈ 3.11
3. Find the Probabilities:
We want to find the probability P(-1.89 < Z < 3.11). This is equal to P(Z < 3.11) - P(Z < -1.89).
Using a z-table or calculator:
P(Z < 3.11) ≈ 0.9991
P(Z < -1.89) ≈ 0.0294
4. Calculate the Probability:
P(-1.89 < Z < 3.11) = 0.9991 - 0.0294 ≈ 0.9697
5. Convert to Percentage:
0.9697 * 100% = 96.97%
Answer:
The probability that a randomly picked fruit will weigh between 251 grams and 346 grams is approximately 96.97%.
