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pablogonzaleztellez
19.02.2020 •
Mathematics
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?
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Ответ:
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.
Ответ:
add the sides that not x to get x
Step-by-step explanation: