To find the probability that the sample mean of a sample size of 16 falls between $28,000 and $32,000, we will use the Central Limit Theorem (CLT). The CLT states that the distribution of the sample means approaches a normal distribution as the sample size (n) increases, provided that the data is drawn from a population with any shape distribution, and the population has a finite mean and standard deviation.
In this problem, we are given:
Population mean (µ) = $30,000
Population standard deviation (σ) = $4,000
Sample size (n) = 16
According to the CLT, the sample mean (x̄) follows a normal distribution with the same mean as the population mean (µ) and a standard deviation equal to the population standard deviation (σ) divided by the square root of the sample size (n):
Mean of x̄ = µ = $30,000
Standard deviation of x̄ = σ/√n = $4,000/√16 = $4,000/4 = $1,000
Now, we want to find the probability that the sample mean falls between $28,000 and $32,000:
P($28,000 ≤ x̄ ≤ $32,000)
To do this, we need to convert the sample mean values into Z-scores using the following formula:
Z = (x̄ - µ) / (σ/√n)
For $28,000:
Z1 = ($28,000 - $30,000) / $1,000 = -2
For $32,000:
Z2 = ($32,000 - $30,000) / $1,000 = 2
Now, we need to find the probabilities corresponding to these Z-scores using a standard normal (Z) table or a calculator with a built-in cumulative distribution function (CDF) for the standard normal distribution:
P(Z ≤ -2) ≈ 0.0228
P(Z ≤ 2) ≈ 0.9772
To find the probability P($28,000 ≤ x̄ ≤ $32,000), we will subtract the smaller probability from the larger one:
P($28,000 ≤ x̄ ≤ $32,000) = P(Z ≤ 2) - P(Z ≤ -2) = 0.9772 - 0.0228 = 0.9544
Thus, the probability that a sample size of 16 has a sample mean between $28,000 and $32,000 is approximately 0.9544, or 95.44%.
