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Burka [1]
1 year ago
6

Average Earnings of Workers The average earnings of year-round full-time workers 25–34 years old with a bachelor’s degree or hig

her 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?

Mathematics
1 answer:
beks73 [17]1 year ago
3 0

Answer:

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.

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Answer: The value of x in trapezoid ABCD is 15

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One of the properties of an isosceles trapezoid is that the angles on either side of the two bases are equal. Since line AD is equal to line BC, then angle D is equal to angle C. It also implies that angle A is equal to angle B.

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A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st
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Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

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<em>Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.</em>

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P(-1.96 < t_2_4_9 < 1.96) = 0.95  {As the critical value of t at 249 degree of

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P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

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                  = [ 15.2-1.96 \times {\frac{2.3}{\sqrt{250} } } , 15.2+1.96 \times {\frac{2.3}{\sqrt{250} } } ]

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<em>The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.</em>

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