A luxury passenger liner has 500 passengers on board whose ages are normally distributed around a mean of 60 years with a standard deviation of 12 years.
3.1 How many of the passengers are between 45 and 78 years old?
3.2 What is the minimum age of the oldest 20% of passengers?
3.1 To find the number of passengers between 45 and 78 years old, we need to find the proportion of passengers within 1 standard deviation below and above the mean:
Z(45) = (45 - 60) / 12 = -1.25
Z(78) = (78 - 60) / 12 = 1.5
Using a standard normal distribution table, we find:
P(Z < -1.25) = 0.1056
P(Z < 1.5) = 0.9332
Therefore, the proportion of passengers between 45 and 78 years old is:
P(-1.25 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.25) = 0.9332 - 0.1056 = 0.8276
Total passengers = 500
Number of passengers between 45 and 78 years old = 0.8276 * 500 = 413.8
Therefore, approximately 414 passengers are between 45 and 78 years old.
3.2 To find the minimum age of the oldest 20% of passengers, we need to find the Z-score associated with the 80th percentile:
Using a standard normal distribution table or calculator, we find the Z-score associated with the 80th percentile is approximately 0.85.
Now we can find the minimum age by solving for x in the equation:
Z = (x - μ) / σ
0.85 = (x - 60) / 12
x - 60 = 0.85 * 12
x = 60 + 10.2
x ≈ 70.2
Therefore, the minimum age of the oldest 20% of passengers is approximately 70.2 years old.