Question

Find the probability that the sample mean starting salary of the 32 randomly selected nurses is less than 69092.4 dollars.

Round final answer to 4 decimal places.

Answer 1

The probability that the sample mean starting salary of 32 nurses is less than $69,092.4 is approximately 0.3944, rounded to four decimal places.

Assuming a normal distribution for starting salaries with a mean (\(μ\)) of $67,694 and a standard deviation σ of $10,333, the Central Limit Theorem is applied for a sample size n of 32.

1. Calculate Parameters:

- Population mean μ: $67,694

- Population standard deviation σ : $10,333

- Sample size n: 32

2. Apply Central Limit Theorem:

- Mean of the sampling distribution (\(μ_¯\)): $67,694

- Standard deviation of the sampling distribution (\(σ_¯\)):
$10,333/√32 ≈ $1,635

3. Calculate Z-Score:

- Z-Score (\(z\)) = \((X_¯ - μ_¯) / σ_¯\)

- For X_¯ = $69,092.4: \(z = (69,092.4 - 67,694) / 1,635 ≈ 0.84\)

4. Find Probability:

- Use a standard normal distribution table or calculator to find the probability associated with z = 0.84.

-
`\(P(\bar{x} < 69,092.4) ≈ 0.3944\)`

5. Final Answer:

- Rounding to four decimal places, the probability that the sample mean starting salary is less than $69,092.4 is approximately 0.3944.