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 less than $290,000. To do this, we can standardize $290,000 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. Plugging in the values we have:

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

Now, we can look up this Z-score in the Z-table to find the probability. The value for Z = 2.0444 is 0.9793. This means that there is a 97.93% chance that the sample mean will be less than $290,000.

So, the probability that the sample mean will be less than $290,000 is 0.9793.

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 less than $290,000. To do this, we can standardize $290,000 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. Plugging in the values we have:

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

Now, we can look up this Z-score in the Z-table to find the probability. The value for Z = 2.0444 is 0.9793. This means that there is a 97.93% chance that the sample mean will be less than $290,000.

So, the probability that the sample mean will be less than $290,000 is 0.9793.

Knowee AI · Accepted Answer

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 less than $290,000. To do this, we can standardize $290,000 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. Plugging in the values we have:

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

Now, we can look up this Z-score in the Z-table to find the probability. The value for Z = 2.0444 is 0.9793. This means that there is a 97.93% chance that the sample mean will be less than $290,000.

So, the probability that the sample mean will be less than $290,000 is 0.9793.

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