Suppose that scores on a particular test are normally distributed with a mean of 110 and a standard deviation of 20. What is the
1 answer:
Answer:
The minimum score needed to be in the top 2% of the scores on the test
is n = 10,00,000
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Mean of the Population = 110
Standard deviation of the Population = 20
The estimated error = 2% = 0.02
<u><em>Step(ii)</em></u>:-
The estimated error is determined by
⇒
Squaring on both sides, we get
n = 10,00,000
The minimum score needed to be in the top 2% of the scores on the test
is n = 10,00,000
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Answer:
The critical value that should be used is T = 2.0796.
The 95% confidence interval for the mean repair cost for the dryers is between $91.912 and $105.648.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 22 - 1 = 21
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 21 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0796, which is the critical value that should be used.
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 98.78 - 6.868 = $91.912
The upper end of the interval is the sample mean added to M. So it is 98.78 + 6.868 = $105.648
The 95% confidence interval for the mean repair cost for the dryers is between $91.912 and $105.648.