Question

college graduates starting salaries a recent survey found that the average starting salaries for college graduates was $49,876. assume that the distribution of starting salaries is approximately normal and the standard deviation is $4329. find these probabilities for randomly selected graduates. use the standard normal distribution table. round the final answer to at least four decimal places and intermediate z value calculations to two decimal places.
part: 0 / 2
part 1 of 2
(a) the graduate makes more than $55,000.
p(x>55,000)=\square

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 49876$, $\sigma=4329$, and $x = 55000$.
$z=\frac{55000 - 49876}{4329}=\frac{5124}{4329}\approx1.18$

Step2: Find the probability

We want to find $P(X>55000)$, which is equivalent to $P(Z > 1.18)$ using the standard normal distribution.
Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard normal distribution table, $P(Z\leq1.18)=0.8810$.
So $P(Z > 1.18)=1 - 0.8810=0.1190$.

Answer:

$0.1190$