SOLUTION: Suppose SPS professors have an average life normally distributed of 73 years, with a population standard deviation of 6 years. a) What percent of SPS professors will live less t

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Question 1191361: Suppose SPS professors have an average life normally distributed of 73 years, with a population standard deviation of 6 years.
a) What percent of SPS professors will live less than 60 years?
b) What proportion of SPS professors will live between 85 and 90 years?
c) Calculate the 30th percentile.
d) Calculate the 97th percentile.
e) What percent of SPS professors will make it past the age of 75?
Answer by CPhill(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to solve this problem:
**a) Percent living less than 60 years:**
1. **Calculate the z-score:**
z = (x - μ) / σ
z = (60 - 73) / 6
z = -2.17
2. **Find the probability:** Use a z-table or calculator to find the area to the *left* of z = -2.17. This gives the probability of living less than 60 years. P(z < -2.17) ≈ 0.015 or 1.5%
**b) Proportion living between 85 and 90 years:**
1. **Calculate the z-scores:**
z₁ = (85 - 73) / 6 = 2
z₂ = (90 - 73) / 6 = 2.83
2. **Find the probabilities:** Use a z-table or calculator.
P(z < 2) ≈ 0.9772
P(z < 2.83) ≈ 0.9977
3. **Find the proportion between the two ages:**
P(2 < z < 2.83) = P(z < 2.83) - P(z < 2) = 0.9977 - 0.9772 ≈ 0.0205
**c) 30th percentile:**
1. **Find the z-score:** The 30th percentile corresponds to a cumulative probability of 0.30. Look up the z-score closest to 0.30 in the z-table; the z-score is approximately -0.52.
2. **Use the z-score formula:**
x = μ + zσ
x = 73 + (-0.52 * 6)
x ≈ 69.88 years
**d) 97th percentile:**
1. **Find the z-score:** The 97th percentile corresponds to a cumulative probability of 0.97. The z-score is approximately 1.88.
2. **Use the z-score formula:**
x = μ + zσ
x = 73 + (1.88 * 6)
x ≈ 84.28 years
**e) Percent making it past 75:**
1. **Calculate the z-score:**
z = (75 - 73) / 6
z = 0.33
2. **Find the probability:** Use a z-table or calculator to find the area to the *right* of z = 0.33. This is 1 - P(z < 0.33). P(z < 0.33) ≈ 0.6293. So, 1 - 0.6293 ≈ 0.3707 or 37.07%