Question

college graduates starting salaries a recent survey found that the average starting salaries for college graduate$49,876. assume that the distribution of starting salaries is approximately normal and the standard deviation is $432these probabilities for randomly selected graduates. use the standard normal distribution table. round the final anto at least four decimal places and intermediate z value calculations to two decimal places.
part 1 of 2
(a) the graduate makes more than $55,000.
$p(x>55,000)=0.1190$
part: 1 / 2
part 2 of 2
(b) the graduate makes between $40,000 and $50,000.
$p(40,000

Explanation:

Step1: Calculate the \(z\) - scores

The formula for the \(z\) - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 49876\) (mean) and \(\sigma = 4320\) (standard deviation).
For \(x = 40000\):
\(z_1=\frac{40000 - 49876}{4320}=\frac{-9876}{4320}\approx - 2.29\)
For \(x = 50000\):
\(z_2=\frac{50000 - 49876}{4320}=\frac{124}{4320}\approx0.03\)

Step2: Use the standard normal distribution table

We know that \(P(40000 < X < 50000)=P(-2.29<Z<0.03)\)
By the property of the standard normal distribution \(P(-2.29 < Z < 0.03)=P(Z < 0.03)-P(Z < - 2.29)\)
From the standard normal distribution table:
\(P(Z < 0.03)=0.5120\)
\(P(Z < - 2.29)=0.0107\)

Step3: Calculate the probability

\(P(-2.29 < Z < 0.03)=0.5120-0.0107 = 0.5013\)

Answer:

\(P(40000 < X < 50000)=0.5013\)