The probability that a randomly selected LU graduate will have a starting salary of at least $30,400 is equal to 96.8%.
Given the following data:
- Standard deviation = $8,000.
- Average starting salary (sample mean) = $20,000.
How to calculate the probability of at least $30,400.
First of all, we would determine the standardized z-score by using this formula:
[tex]Z=\frac{\bar{x}\;-\;\mu}{ \sigma }\\\\Z=\frac{30400\;-\;20000}{8000}\\\\Z=\frac{10400}{8000 }[/tex]
Z = 1.30.
From the z-table, the p-value is given by:
P(X ≥ 30400) = P(X > 30400)
P(X > 30400) = P(Z > 1.30)
P(Z > 1.30) = 1 - P(Z < 1.30)
P(Z > 1.30) = 1 - 0.9032
P(Z > 1.30) = 0.0968.
P(X ≥ 30400) = 0.0968 = 96.8%.
For P(Z < 15600), we have:
[tex]Z=\frac{\bar{x}\;-\;\mu}{ \sigma }\\\\Z=\frac{15600\;-\;20000}{8000}\\\\Z=\frac{-4400}{8000 }[/tex]
Z = -0.55.
P(Z < 15600) = P(Z < -0.55)
P(Z < -0.55) = 0.2912 = 29.12%.
How to calculate the starting salaries of the middle 95%.
In order to determine the minimum and maximum starting salaries of the middle 95%, we would find the range by using this formula:
[tex]R = \mu \pm (Z_{\alpha/2} )\sigma[/tex]
For alpha using the z-table, we have:
α = 1 - 0.95
α = 0.05.
[tex]Z_{\alpha/2} = Z_{0.05/2}\\\\Z_{0.025}=1.96[/tex]
Now, we can calculate the minimum (lower) starting salaries:
Minimum = 20000 - 1.96(8000)
Minimum = 20,000 - 15,680
Minimum = $4,320.
For the maximum (upper):
Maximum = 20000 + 1.96(8000)
Maximum = 20,000 + 15,680
Maximum = $35,680.
Read more on standard deviation here: https://brainly.com/question/4302527
