The incomes in a certain lg. population of college teachers have a normal distribution with a mean of $35,000 and st. dev. of $5000. Four teachers are randomly selected. What is the probability their average salary will exceed $40,000?
mean = 35000
St.dev =5000
x = 40000
z = (x - mean)/sd/sqrt(n)
n =
z = (40000-35000)/5000/sqrt(4)
z = 5000/(5000/2)
z = 2
P(z >2)
mean = 35000
st.dev = 5000
x = 40000
n = 4
z = (40000-35000)/5000/sqrt(4)
z = 5000/(5000/2)
z = 2
P(z >2) = 0.0228
To find the probability that the average salary of four randomly selected teachers will exceed $40,000, we need to find the corresponding z-score and then look up the probability from a standard normal distribution table.
Step 1: Find the z-score.
The z-score formula is given by:
z = (x - μ) / σ
Where:
x = $40,000 (desired average salary)
μ = $35,000 (mean salary)
σ = $5,000 (standard deviation)
Substituting the values into the formula, we get:
z = (40,000 - 35,000) / 5,000
Calculating this, we have:
z = 5,000 / 5,000
z = 1
Step 2: Use the z-score to find the probability.
We need to find the probability that the z-score is greater than 1. This corresponds to finding the area under the standard normal distribution curve to the right of the z-score.
Using a standard normal distribution table or calculator, we can find that the probability of obtaining a z-score greater than 1 is approximately 0.1587.
Therefore, the probability that the average salary of four randomly selected teachers will exceed $40,000 is approximately 0.1587 or 15.87%.