Suppose that the population of weights of airline passengers has mean 150 lbs. and standard deviation 25 lbs. A certain plane has a capacity of 7,800 lbs.
(a) What is the probability that a flight of 50 passengers will exceed capacity it?
(b) If we wish to reduce the chance of overload to 0.01, how much must the capacity be increased?
To answer these questions, we need to use the concept of the normal distribution and z-scores.
(a) To find the probability that a flight of 50 passengers will exceed capacity, we need to calculate the z-score for the capacity limit using the given mean and standard deviation. Then, we can use a standard normal distribution table or a calculator to find the probability.
1. Calculate the standard deviation of the sample mean (also known as the standard error):
Standard Deviation of Sample Mean (SE) = Standard Deviation of Population / Square Root of Sample Size
SE = 25 / √(50)
2. Calculate the z-score for the capacity limit:
z = (Capacity Limit - Mean) / SE
z = (7800 - 150) / (25 / √50)
3. Use the z-score to find the probability of exceeding the capacity using a standard normal distribution table or a calculator.
(b) To determine how much the capacity must be increased to reduce the chance of overload to 0.01, we need to find the corresponding z-score for a 0.01 probability. Then, we can use the same formula as above to calculate the increased capacity.
1. Find the z-score for a 0.01 probability using a standard normal distribution table or a calculator.
2. Solve the equation for the increased capacity:
New Capacity = (z-score * SE) + Mean
I'll perform the calculations and provide the answers for you.