Question

college graduates starting salaries a recent survey found that the average starting salaries for college graduates was $49,256. assume that the distribution of starting salaries is approximately normal and the standard deviation is $4791. 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 1 of 2 (a) the graduate makes more than $55,000. p(x>55,000)=0.1151 part: 1 / 2 part 2 of 2 (b) the graduate makes between $40,000 and $50,000. p(40,000<x<50,000)=

Explanation:

Step1: Calculate the $z$-scores

The formula for the $z$-score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 49256$ (mean) and $\sigma=4791$ (standard deviation).
For $x = 40000$:
$$z_1=\frac{40000 - 49256}{4791}=\frac{-9256}{4791}\approx - 1.93$$
For $x = 50000$:
$$z_2=\frac{50000 - 49256}{4791}=\frac{744}{4791}\approx0.16$$

Step2: Use the standard - normal distribution table

We know that $P(40000 < X < 50000)=P(-1.93By the property of the normal distribution $P(a < Z < b)=P(Z < b)-P(Z < a)$.
From the standard - normal distribution table, $P(Z < 0.16)=0.5636$ and $P(Z<-1.93) = 0.0247$.

Step3: Calculate the probability

$$P(-1.93 < Z < 0.16)=P(Z < 0.16)-P(Z < - 1.93)=0.5636-0.0247 = 0.5389$$

Answer:

$P(40000 < X < 50000)=0.5389$