Question

Random samples of 576 are taken from a large population and studied. It was found that σ

x
ˉ

​
=9.31. If 12.3% of all sample means were greater than 269.3996. What is μ ? μ=

Answer 1

The value of μ is approximately 258.6

Step-by-step explanation:

To find the value of μ, we need to use the formula for the standard error of the mean (σxˉ):

σxˉ = σ / √n

Where σ is the population standard deviation and n is the sample size.

Given that σxˉ = 9.31 and n = 576, we can rearrange the formula to solve for σ:

σ = σxˉ * √n

σ = 9.31 * √576

σ = 9.31 * 24

σ = 223.44

Now, we can use the Z-score formula to find the Z-score corresponding to the given percentage:

Z = (x - μ) / (σ / √n)

In this case, x is 269.3996, and the percentage is 12.3%. We need to find the Z-score that corresponds to the upper tail of the distribution, so we subtract the percentage from 100%:

Z = (269.3996 - μ) / (223.44 / √576)

To find the Z-score corresponding to 100% - 12.3% = 87.7%, we can use a Z-table or a calculator. The Z-score that corresponds to 87.7% is approximately 1.16.

1.16 = (269.3996 - μ) / (223.44 / √576)

Now, we can solve for μ:

1.16 * (223.44 / √576) = 269.3996 - μ

(1.16 * 223.44) / √576 = 269.3996 - μ

(1.16 * 223.44) / √576 + μ = 269.3996

μ = 269.3996 - (1.16 * 223.44) / √576

μ ≈ 258.6

Therefore, the value of μ is approximately 258.6.

Answer 2

Random samples of 576 are taken from a large population and studied. It was found that is μ = 267.56

Step-by-step explanation:

To find the population mean (μ), we can use the formula for the standard error of the mean (σxˉ​) which is given by σxˉ​ = σ / √n, where σ is the population standard deviation and n is the sample size. In this case, we are given that σxˉ​ = 9.31 and n = 576. Rearranging the formula to solve for σ, we get σ = σxˉ​ √n = 9.31 √576 = 9.31 * 24 = 223.44.

Next, we can use the z-score formula to find the z-score corresponding to the given percentage of sample means greater than a certain value.

The z-score formula is z = (x - μ) / (σ / √n), where x is the value, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Given that 12.3% of all sample means were greater than 269.3996, we can calculate the z-score using this information.

Using a standard normal distribution table or calculator, we find that the z-score corresponding to 12.3% is approximately 1.17.

Now, we can use this z-score to solve for μ in the z-score formula. Rearranging the formula to solve for μ, we get μ = x - (z (σ / √n)) = 269.3996 - (1.17 (223.44 / √576)) ≈ 267.56.

Therefore, the population mean (μ) is approximately 267.56.