Final answer:
To find the probability, we need to calculate the z-score for the sample mean of 23.5 lb using the formula: z = (X - μ) / (σ / sqrt(n)). Using a standard normal distribution table or a calculator, we can find that the probability corresponding to this z-score is approximately 0.0228.
Step-by-step explanation:
To find the probability that the passengers will have a mean weight under 23.5 lb, we need to use the concept of sampling distribution of the mean. The mean weight of baggage for passengers on airline A follows a normal distribution with a mean (μ) of 25 lb and a standard deviation (σ) of 5 lb. Since we have a sample size of 45, we can use the central limit theorem.
The central limit theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution. In this case, we can approximate the distribution of the sample mean with a normal distribution.
To find the probability, we need to calculate the z-score for the sample mean of 23.5 lb using the formula:
z = (X - μ) / (σ / sqrt(n))
Where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Plugging in the values, we get:
z = (23.5 - 25) / (5 / sqrt(45))
z = -1.5 / (5 / 6.71)
z = -1.5 / 0.747
z = -2.008
Using a standard normal distribution table or a calculator, we can find that the probability corresponding to this z-score is approximately 0.0228.
Therefore, the probability that the passengers will have a mean weight under 23.5 lb is approximately 0.0228.
The complete question...............The baggage weights for passengers on airline A have a mean (μ) of 25 lb and a standard deviation of 5 lb. If there are 45 people on a flight, find the probability that the passengers will have a mean under 23.5 lb.
a) 0.352
b) 0.648
c) 0.815
d) 0.952.................../;