word math

Question

According to data released by the Changer of Commerce in a certain city, the weekly wages of factory workers are normally distributed with a mean of $720 and a standard deviation of $60.  What is the probability that a factory worker selected at random from the city make a weekly wage:

 

a.  Of less than $720?

b.  Of more than $912?

c.  Between $660 and $780?

Answer 1

a)  Since $720 is the mean, then the probability of making less than the mean is the same as making more than the mean, so that's 0.50

 

b)  We need a z-score for this, so z = (912 - 720)/60 = 3.2

If you go to the +z-score table and look under 3.20, it says p = 0.9993. That's the probability of making less than $912. So making more than $912 = 1 - 0.9993 = 0.0007

 

c)  z-score for 660 = (660 - 720)/60 = -1.00, and z-score for 780 = +1. So the known probability between ±1 standard deviation is 0.68. (In other words, 68% of workers make between these 2 dollar amounts, or there's a 68% chance that a randomly selected worker makes between these 2 dollar amounts).

 

By the way, it's CHAMBER of Commerce.