Average Earnings of Workers The average earnings of year-round full-time workers 25–34 years old with a bachelor’s degree or higher were $58,500 in 2003. If the standard deviation is $11,200, what can you say about the percentage of these workers who earn?
a. Between $47.300 and $69,700?
b. More than $80.900?
c. How likely is it that someone earns more than $100,000?

a. 68% of the workers will earn between $47300 and $69700.

b. 2.5% of workers will earn above $89000

c. Approximately 0

Step-by-step explanation:

The standard normal distribution curve in the attached graph is used to solve this question.

a. The value $47300 is a standard deviation below the mean i.e. 58500-11200=47300. While $69700 is a standard deviation above the mean. I.e. 58500+12000=69700.

Between the first deviation below and above the mean, you have 34%+34%=68% of the salary earners within this range. So we have 68%of staffs earning within this range

b. The second standard deviation above the mean is $80900. i.e. 58500+11200+11200=$80900

We have 50%+13.5%+2.5%= 97.5% earning below $80900. Therefore, 100-97.5= 2.5% of the workers earn above this amount.

c. From the Standard Deviation Rule, the probability is only about (1 -0 .997) / 2 = 0.0015 that a normal value would be more than 3 standard deviations away from its mean in one direction or the other. The probability is only 0.0002 that a normal variable would be more than 3.5 standard deviations above its mean. Any more standard deviations than that, and we generally say the probability is approximately zero.

(a) the volume of the cone is 20/3 in3

hence, option a is correct.