In a certain region, the mean annual salary for plumbers is $51,000. Let x be a random variable that represents a plumber's salary. Assume the standard deviation is $1300. If a random sample of 100 plumbers is selected, what is the probability that the sample mean is greater than $51,300?

The population mean annual salary for plumber is 46,700. A random sample of 42 plumber is drawn from this population. What is the probality that the mean salary of the sample is greater than 44,000? Assume that standard deviation is 5,600

0.00090

Use z-scores.

Formula for this problem:

z = (x - mean)/(sd/√n)

With your data:

z = (51300 - 51000)/(1300/√100)

I'll let you finish the calculation.

Once you find the z-score, use a z-table to determine your probability. (Remember that the problem is asking for the probability that the sample mean is "greater than" 51300 meters.)

.31

To solve this problem, we can use the Central Limit Theorem. According to the Central Limit Theorem, if the sample size is large enough (typically n > 30), the distribution of the sample mean tends to be approximately normal regardless of the shape of the population distribution.

First, we need to find the z-score for the given sample mean of $51,300. The formula to calculate the z-score is:

z = (x - μ) / (σ / √n)

Where:
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size

In this case:
x = $51,300
μ = $51,000
σ = $1300
n = 100

Substituting the values into the formula:

z = (51,300 - 51,000) / (1300 / √100)
z = 300 / (1300 / 10)
z = 300 / 130
z ≈ 2.31 (rounded to two decimal places)

Next, we need to find the corresponding probability using a standard normal distribution table or calculator. The probability corresponds to the area under the curve to the right of the z-score.

Using a standard normal distribution table or calculator, we find that the probability of obtaining a z-score greater than 2.31 is approximately 0.0104.

Therefore, the probability that the sample mean is greater than $51,300 is approximately 0.0104 or 1.04%.