Answer to Question #327303 in Statistics and Probability for Tarek

Question

The weekly income of managers are normally distributed with a mean of $10000.

And a standard deviation of 100S.

1: What is the z value for an income x of 1100 and for 900?

2. What is the probability that the weekly income is less than 1100?

3. What is the probability that the weekly income is more than 1100?

4. What is the probability that the weekly income between 800S and 1100$?

Answer 1

$P(900<X<1100)=P(\frac{900-10000}{100}<Z<\frac{1100-10000}{100})=P(-91<Z<-89)=0$

2. $P(X<1100)=P(Z<\frac{1100-10000}{100})=P(Z<-89)=0$

$P(800<X<1100)=P(\frac{800-10000}{100}<Z<\frac{1100-10000}{100})=P(-92<Z<-89)=0$