The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 inches. Heights of men are normally distributed with a mean of 69 inches and a standard deviation of 2.8 inches.
b) assume that half of the 200 passengers are men, what doorway height satisfies the condition that there is a .95 probability that this height is greater than the mean height of 100 men?
0.8577
To find the doorway height that satisfies the condition, we need to use the concept of standard deviation and normal distribution.
Step 1: Calculate the mean height of the 100 men:
The mean height of the men is given as 69 inches.
Step 2: Calculate the standard deviation of the mean height of the 100 men:
The standard deviation of the men's height is given as 2.8 inches. Since we are considering the mean height of 100 men, we divide the standard deviation by the square root of the sample size (100) to get the standard deviation of the mean.
Standard deviation of the mean = 2.8 / √100 = 2.8 / 10 = 0.28 inches
Step 3: Convert the desired probability of the height being greater than the mean height of the 100 men to a Z-score:
Given that we want a 0.95 probability, we need to find the Z-score that corresponds to this probability. We can use a standard normal distribution table or calculator to find the Z-score. For a probability of 0.95, the Z-score is approximately 1.645.
Step 4: Calculate the doorway height that satisfies the condition:
To find the doorway height, we need to add the Z-score times the standard deviation of the mean to the mean height of the 100 men.
Doorway height = Mean height of 100 men + (Z-score * Standard deviation of the mean)
Doorway height = 69 + (1.645 * 0.28)
Doorway height = 69 + 0.4616
Doorway height ≈ 69.4616 inches
Therefore, the doorway height that satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men is approximately 69.4616 inches.
To find the height that satisfies the given condition, we need to calculate the z-score and then find the corresponding height from the standard normal distribution.
First, let's find the z-score for the given condition. The z-score formula is given by:
z = (x - μ) / σ
where:
x = height we want to find
μ = mean height of men (69 inches)
σ = standard deviation of men's heights (2.8 inches)
To find the height (x) such that there is a 0.95 probability that it is greater than the mean height of 100 men, we need to find the z-score that corresponds to a 0.95 probability (1 - 0.05).
Using a standard normal distribution table or a calculator, we can find the z-score corresponding to a 0.95 probability to be approximately 1.645.
Now we can rearrange the z-score formula to solve for the height (x):
x = z * σ + μ
Substituting the values we have:
x = 1.645 * 2.8 + 69
Calculating:
x ≈ 4.606 + 69
x ≈ 73.606
Therefore, the doorway height that satisfies the condition is approximately 73.606 inches.
a)
The value of z such that P(Z<z)= 0.95 is z = 1.6449
Solving for x:
z =1.6449=(x-69.0)/2.8
x = 1.6449*2.8+69 = 73.6
b)
Solving for xbar:
z =1.6449=(xbar-69.0)/(2.8/sqrt(100))
xbar = 1.6449*0.28+69 = 69.5