The mean number of bumped airline passengers per day is 1.35 and the standard deviation is 0.25. For a random selection of a group of 40 days, what is the probability that the mean of bumped passengers for the 40 days will be between 1.25 and 1.50?
Use this z-score formula for this problem:
z = (x - mean)/(sd/√n)
x = 1.25, 1.50
mean = 1.35
sd = 0.25
n = 40
Calculate two z-scores, then use a z-table to determine probability between the two scores.
I hope this will help get you started.
z=(1.25-1.35)/(0.25sqrt40)= -0.0791
z=(1.50-1.35)/(0325sqrt40)= 0.0791
but not sure how to read the z scores I don't know if it would be.5000-.5000 or what can someone please help me on the z scores for -0.0791 and 0.0791
SD/√n = .25/6.3246 = .039528192 = .04
Z = (1.25-1.35)/.04 = -.10/.04 = -2.5
Z = (1.50-1.35)/.04 = .15/.04 = ?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of both Z scores between those Z scores and mean. Add the two together.
To find the probability that the mean of bumped passengers for 40 days will be between 1.25 and 1.50, we can use the Central Limit Theorem and the standard normal distribution.
According to the Central Limit Theorem, for a large sample size (n > 30 in this case), the sampling distribution of the sample means will be approximately normal, regardless of the shape of the population distribution.
First, let's find the standard error (SE) of the mean for the 40-day sample. The standard error is the standard deviation of the population divided by the square root of the sample size.
SE = standard deviation / √sample size
SE = 0.25 / √40
SE ≈ 0.0397
Next, we need to standardize the sample mean using the z-score formula:
z = (x - μ) / (SE)
where x is the value we are interested in (1.25 and 1.50 in this case), μ is the population mean (1.35 in this case), and SE is the standard error.
For 1.25:
z1 = (1.25 - 1.35) / 0.0397 ≈ -2.52
For 1.50:
z2 = (1.50 - 1.35) / 0.0397 ≈ 3.78
Now, we can use the standard normal distribution table or a calculator to find the probability of getting a z-score between -2.52 and 3.78.
P(-2.52 < z < 3.78) ≈ P(z < 3.78) - P(z < -2.52)
Using a standard normal distribution table, we can find the probabilities: P(z < 3.78) ≈ 1 and P(z < -2.52) ≈ 0.0056.
Therefore, the probability that the mean of bumped passengers for the 40 days will be between 1.25 and 1.50 is approximately 1 - 0.0056 = 0.9944, or 99.44%.