Answer :
Answer:
A. 0.0021
Step-by-step explanation:
Given that the heights of men are normally distributed with a mean of 66.9 inches and a standard deviation of 2.1inches.
Sample size = 36
Std dev of sample = [tex]\frac{2.1}{\sqrt{36} } =0.35[/tex]
The sample entries X the heights are normal with mean= 66.9 inches and std deviation = 0.35 inches
Or we have
Z = [tex]\frac{x-66.9}{0.35}[/tex]
Hence the probability that they have a mean height greater than 67.9 inches
=[tex]P(X>67.9)\\=P(Z>\frac{1}{0.35)} \\=0.00214[/tex]
So option A is right answer.
Final answer:
To find the probability that the mean height of 36 randomly selected men is greater than 67.9 inches, calculate the z-score and find the corresponding area under the standard normal distribution curve.
Explanation:
To find the probability that the mean height of 36 randomly selected men is greater than 67.9 inches, we need to calculate the z-score and find the corresponding area under the standard normal distribution curve.
The z-score is calculated using the formula:
z = (x - μ) / (σ / √n)
Where x is the value we want to find the probability for, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Substituting the given values into the formula, we have:
z = (67.9 - 66.9) / (2.1 / √36) = 1.71
Using a standard normal distribution table or a calculator, we can find that the area to the right of a z-score of 1.71 is approximately 0.0436.
Since we want the probability of having a mean height greater than 67.9 inches, we need to calculate the area to the right of the z-score. Therefore, the probability is approximately 1 - 0.0436 = 0.9564.
