Suppose that the height of male students at a large university is normally distributed with a mean of 69 inches and a standard deviation of 3.0 inches. If 16 samples consisting of 25 students each are obtained, what is the probability that the mean of these 16 samples is between 68 and 70 inches?

Z = (mean1 - mean2)/standard error of the mean (SEm)

SEm = SD/√n(in each sample)

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to these Z scores.

0.818

To find the probability that the mean of the 16 samples is between 68 and 70 inches, we can use the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size increases.

First, we need to calculate the standard deviation of the sample distribution, also known as the standard error. The standard error is given by:

SE = σ / √n

Where σ is the standard deviation of the population (3.0 inches) and n is the sample size (25 students).

SE = 3.0 / √25
SE = 3.0 / 5
SE = 0.6 inches

Next, we need to calculate the z-scores for the lower and upper limits of the desired range.

Lower z-score:
z1 = (x1 - μ) / SE = (68 - 69) / 0.6 = -1.67

Upper z-score:
z2 = (x2 - μ) / SE = (70 - 69) / 0.6 = 1.67

Now, we can use a standard normal distribution table or a calculator to find the area under the curve between these z-scores. We'll need to find the cumulative probability between z1 and z2.

P(z1 < Z < z2) = P(-1.67 < Z < 1.67)

Using a standard normal distribution table or calculator, we find that the area between z = -1.67 and z = 1.67 is approximately 0.8920.

Therefore, the probability that the mean of these 16 samples is between 68 and 70 inches is 0.8920, or 89.20%.

To solve this problem, we need to use the Central Limit Theorem, which states that the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is sufficiently large.

In this case, we have 16 samples, each consisting of 25 students. The sample mean is the average height of each sample.

Given that the population mean is 69 inches and the standard deviation is 3.0 inches, we can find the standard deviation of the distribution of sample means, also known as the standard error, by dividing the population standard deviation by the square root of the sample size:

Standard error = population standard deviation / square root of sample size
Standard error = 3.0 / sqrt(25) = 3.0 / 5 = 0.6 inches

Now, we can convert the interval [68, 70] to z-scores using the formula:

z = (x - μ) / σ

where z is the z-score, x is the value we want to convert, μ is the population mean, and σ is the standard deviation.

For the lower bound:
z1 = (68 - 69) / 0.6 = -1.67

For the upper bound:
z2 = (70 - 69) / 0.6 = 1.67

Now, we need to find the probability that the mean of these 16 samples falls between the two z-scores. This can be done by finding the area under the normal distribution curve between the two z-scores.

We can use a standard normal distribution table or a calculator to find these probabilities. Alternatively, we can use the cumulative distribution function (CDF) of the standard normal distribution.

The probability can be calculated as follows:

P(-1.67 < Z < 1.67) = P(Z < 1.67) - P(Z < -1.67)

Using a standard normal distribution table or a calculator, we find:

P(Z < 1.67) = 0.9525
P(Z < -1.67) = 1 - P(Z < 1.67) = 1 - 0.9525 = 0.0475

Therefore:

P(-1.67 < Z < 1.67) = 0.9525 - 0.0475 = 0.905

So, the probability that the mean of these 16 samples is between 68 and 70 inches is 0.905, or 90.5%.