Sampling Distribution
Complete solutions and illustrations are required.
1.The average annual salary for all U.S. teachers is $29,863. Assume that the distribution is normal and the standard deviation is $5,200.
a.Find the probability that a randomly selected teacher earns at least $35,000 a year.
b.What is the probability that the mean for a sample of 80 teachers’ salaries is greater than $30,000?
To find the probabilities in these questions, we need to use the concept of the sampling distribution. The sampling distribution refers to the distribution of sample means taken from a population. In this case, we will be using the normal distribution to approximate the sampling distribution.
1a. To find the probability that a randomly selected teacher earns at least $35,000 a year, we need to standardize the value and then look up the corresponding z-score in the standard normal distribution table.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
Where:
x = $35,000 (the value we are interested in)
μ = $29,863 (mean of the population)
σ = $5,200 (standard deviation of the population)
z = (35,000 - 29,863) / 5,200
z = 1.00
Now we need to find the probability corresponding to a z-score of 1.00 in the standard normal distribution table. Looking it up, we find that the probability is approximately 0.8413.
Therefore, the probability that a randomly selected teacher earns at least $35,000 a year is approximately 0.8413 or 84.13%.
1b. To find the probability that the mean for a sample of 80 teachers' salaries is greater than $30,000, we need to calculate the z-score for the sample mean.
The mean of the sampling distribution will be the same as the population mean, μ = $29,863. However, the standard deviation of the sampling distribution (also called the standard error) is calculated using the formula:
σᵡ = σ / sqrt(n)
Where:
σ = $5,200 (standard deviation of the population)
n = sample size = 80
σᵡ = 5,200 / sqrt(80)
σᵡ ≈ 582.91
Now we calculate the z-score:
z = (x - μ) / σᵡ
Where:
x = $30,000 (the value we are interested in)
μ = $29,863 (mean of the population)
σᵡ ≈ 582.91 (standard error)
z = (30,000 - 29,863) / 582.91
z ≈ 0.235
We need to find the probability corresponding to a z-score of 0.235 in the standard normal distribution table. Looking it up, we find that the probability is approximately 0.5918.
Therefore, the probability that the mean for a sample of 80 teachers' salaries is greater than $30,000 is approximately 0.5918 or 59.18%.