Suppose x has a distributionwith u=72 and 0=8 if random samples of size n=16 are selected, can we say anything about the x distributin of sample means

if the originalx distribution is normal can we say anything about the x distribution of random samples of size 16 find p(68<x,< 73

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

If only one SD is provided, you can use just that to determine SEdiff.

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

If the original x distribution is normal with mean μ = 72 and standard deviation σ = 8, then the sample means will also be normally distributed. For the distribution of sample means, the mean (μ') will be equal to the population mean (μ) and the standard deviation (σ') will be equal to the population standard deviation (σ) divided by the square root of the sample size (n).

In this case, the sample size is n = 16, so the standard deviation of the sample means (σ') will be σ/√n = 8/√16 = 2.

To find the probability that a sample mean (x̄) falls between 68 and 73, we need to calculate the z-scores corresponding to those values and then find the probability using the standard normal distribution.

The z-score for 68 can be calculated as:

Z1 = (68 - μ') / σ' = (68 - 72) / 2 = -2

The z-score for 73 can be calculated as:

Z2 = (73 - μ') / σ' = (73 - 72) / 2 = 0.5

Now, we need to find the probability that a z-score falls between -2 and 0.5. We can use a standard normal distribution table (z-table) to find these probabilities.

Using the z-table, we find that the probability corresponding to a z-score of -2 is 0.0228, and the probability corresponding to a z-score of 0.5 is 0.6915.

To find the probability that a sample mean falls between 68 and 73 (P(68 < x̄ < 73)), we subtract the probability corresponding to -2 from the probability corresponding to 0.5:

P(68 < x̄ < 73) = 0.6915 - 0.0228 = 0.6687

Therefore, the probability that a random sample mean falls between 68 and 73 is approximately 0.6687, assuming the original x distribution is normal.

To understand if we can say anything about the distribution of sample means, we need to consider the central limit theorem (CLT). The central limit theorem states that, regardless of the distribution of the original variable as long as the sample size is sufficiently large (usually n > 30), the distribution of the sample means will approach a normal distribution.

In your case, since the sample size is 16, which is smaller than the generally recommended threshold for the CLT, we cannot directly assume that the distribution of the sample means will be normal. However, it is worth noting that the normal approximation may still hold reasonably well for practical purposes if the population distribution is approximately symmetric and not heavily skewed.

To answer your second question, if we assume that the original distribution of x is normal, then we can have some insights into the distribution of random samples of size 16. Specifically, the sampling distribution of the sample means will also be normal, regardless of the sample size, due to the properties of the normal distribution.

Now, to find the probability P(68 < x < 73) for the distribution of sample means, we need to convert the values to z-scores and use the standard normal distribution table (also known as the z-table).

First, we need to calculate the z-scores for 68 and 73 using the formula: z = (x - μ) / σ, where μ is the mean of the original distribution and σ is the standard deviation. In this case, μ = 72 and σ = 8.

For 68: z1 = (68 - 72) / 8 = -0.5
For 73: z2 = (73 - 72) / 8 = 0.125

Next, we can look up the corresponding probabilities in the standard normal distribution table for these z-scores. The z-table provides the cumulative probability up to a given z-score.

Let's assume the z-table shows that the cumulative probability for z1 is 0.3085 and the cumulative probability for z2 is 0.5508.

To find the probability between 68 and 73, we subtract the probability of z1 from the probability of z2:
P(68 < x < 73) = P(z1 < Z < z2) = P(z2) - P(z1) = 0.5508 - 0.3085 = 0.2423

Therefore, the probability that a random sample mean falls between 68 and 73, assuming the original distribution is normal, is approximately 0.2423.