Recall that the population average of the heights in the file "pop1.csv" is μ =
170.035. Using simulations we found that the probability of the sample average of the height falling within 1 centimeter of the population
...
Recall that the population average of the heights in the file "pop1.csv" is μ =
170.035. Using simulations we found that the probability of the sample average of the height falling within 1 centimeter of the population average is approximately equal to 0.626. From the simulations we also got that the standard deviation of the sample average is (approximately) equal to
1.122. In the next 3 questions you are asked to apply the Normal approximation to the distribution of the sample average using this information. The answer may be rounded up to 3 decimal places of the actual value:
Using the Normal approximation, the probability that sample average of the heights falls within 1 centimeter of the population average is
Answer: 0.6272133
The correct answer is: 0.6272133
Using the Normal approximation we get that the central region that contains 95% of the distribution of the sample average is of the form
170.035 ± z · 1.122. The value of z is
Answer: 1.959964
Using the Normal approximation, the probability that sample average of the heights is less than 168 is
Answer: 0.03486
The correct answer is: 0.03486
According to the Internal Revenue Service, the average length of time for an individual to complete (record keep, learn, prepare, copy, assemble and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is 2 hours. Suppose we randomly sample 36 taxpayers and compute their average time to completing the forms. Then the probability that the average is more than 11 hours is approximately equal to (The answer may be rounded up to 3 decimal places of the actual value.) 0.07926 .
The correct answer is: 0.07926984
Suppose that a category of world class runners are known to run a marathon (26 miles) in an expectation of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. In the next 3 questions you are asked to apply the Normal approximation to the distribution of the sample average using this information. The answer may be rounded up to 3 decimal places of the actual value:
The probability that the runner will average between 142 and 146 minutes in these 49 marathons is
[Show More]