The heights of 18-year-old men are approximately normally distributed with mean 68 inches and standard deviation 3 inches. What is the probability that the average height of a sample of ten 18-year-old men will be less than 70 inches? Round your answer to four decimal places.

0.0174
0.4826
0.9652
0.9826
0.4913

Are you looking for the average or the individual score? This would be for the latter. For the former, you need to have the sample size to find the SE (see following post).

Z = (score - mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.

Sorry, it was not your post. Here it is.

98% interval = mean ± SE

SE (Standard error) = SD/√(n-1)

Z = (score - mean)/SE

To solve this problem, we need to use the Central Limit Theorem, which states that the distribution of sample means will be approximately normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

Given:
Population mean (μ) = 68 inches
Population standard deviation (σ) = 3 inches
Sample size (n) = 10

We need to find the probability that the average height of a sample of ten 18-year-old men will be less than 70 inches.

First, we need to standardize the sample mean using the formula for a standard normal distribution:

Z = (x - μ) / (σ / sqrt(n))

Where:
x = sample mean (70 inches)
μ = population mean (68 inches)
σ = population standard deviation (3 inches)
n = sample size (10)

Plugging in the given values into the formula:

Z = (70 - 68) / (3 / sqrt(10))
Z = 2 / (3 / sqrt(10))
Z = 2 * sqrt(10) / 3

Now, we need to find the probability associated with a Z-value of 2 * sqrt(10) / 3. We can look up this probability in a standard normal distribution table or use a calculator:

P(Z < 2 * sqrt(10) / 3) ≈ 0.9826

Therefore, the probability that the average height of a sample of ten 18-year-old men will be less than 70 inches is approximately 0.9826, rounded to four decimal places.

The correct answer is 0.9826.

To solve this problem, we can use the Central Limit Theorem. According to the Central Limit Theorem, when the sample size is sufficiently large, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution.

In this case, we are given that the heights of 18-year-old men are normally distributed with a mean of 68 inches and a standard deviation of 3 inches. We want to find the probability that the average height of a sample of ten 18-year-old men will be less than 70 inches.

To find this probability, we need to first standardize the sample mean using the formula for the standard deviation of the sample mean:

Standard deviation of the sample mean = Standard deviation of the population / Square root of the sample size

In this case, the standard deviation of the population is 3 inches, and the sample size is 10. Therefore, the standard deviation of the sample mean is 3 / sqrt(10) = 0.9487 (rounded to four decimal places).

Next, we need to calculate the z-score, which represents how many standard deviations the sample mean is away from the population mean:

z = (sample mean - population mean) / standard deviation of the sample mean
= (70 - 68) / 0.9487
= 2.109 (rounded to three decimal places)

Finally, we can use a standard normal distribution table or a calculator to find the probability corresponding to the z-score of 2.109. From the table or calculator, we find that the probability is 0.9826 (rounded to four decimal places).

Therefore, the correct answer is: 0.9826.