To solve this problem, we can use the Central Limit Theorem which states that if we have a sufficiently large sample size, the sampling distribution of the mean will be approximately normally distributed.

The mean (μ) of the sampling distribution of the mean is equal to the population mean, which is $271,600.

The standard deviation (σ) of the sampling distribution of the mean, also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size. So, σ = $90,000 / sqrt(100) = $9,000.

We want to find the probability that the sample mean is between $280,000 and $290,000. To do this, we can standardize these values using the Z-score formula:

Z = (X - μ) / σ

Where X is the value we are interested in, μ is the mean, and σ is the standard deviation.

For $280,000:

Z1 = ($280,000 - $271,600) / $9,000 = 0.9333

For $290,000:

Z2 = ($290,000 - $271,600) / $9,000 = 2.0444

Now, we can look up these Z-scores in the Z-table to find the probabilities. The value for Z1 = 0.9333 is 0.8249 and for Z2 = 2.0444 is 0.9793.

The probability that the sample mean will be between $280,000 and $290,000 is the difference between these two probabilities: 0.9793 - 0.8249 = 0.1544.

So, the probability that the sample mean will be between $280,000 and $290,000 is 0.1544.

Question

To solve this problem, we can use the Central Limit Theorem which states that if we have a sufficiently large sample size, the sampling distribution of the mean will be approximately normally distributed.

The mean (μ) of the sampling distribution of the mean is equal to the population mean, which is $271,600.

The standard deviation (σ) of the sampling distribution of the mean, also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size. So, σ = $90,000 / sqrt(100) = $9,000.

We want to find the probability that the sample mean is between $280,000 and $290,000. To do this, we can standardize these values using the Z-score formula:

Z = (X - μ) / σ

Where X is the value we are interested in, μ is the mean, and σ is the standard deviation.

For $280,000:

Z1 = ($280,000 - $271,600) / $9,000 = 0.9333

For $290,000:

Z2 = ($290,000 - $271,600) / $9,000 = 2.0444

Now, we can look up these Z-scores in the Z-table to find the probabilities. The value for Z1 = 0.9333 is 0.8249 and for Z2 = 2.0444 is 0.9793.

The probability that the sample mean will be between $280,000 and $290,000 is the difference between these two probabilities: 0.9793 - 0.8249 = 0.1544.

So, the probability that the sample mean will be between $280,000 and $290,000 is 0.1544.

Knowee AI · Accepted Answer