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antoniya [11.8K]
3 years ago
5

The University of Metropolis requires its students to pass an examination in college-level mathematics before they can graduate.

The students are given three chances to pass the exam; 61% pass it on their first attempt, 63% of those that take it a second time pass it then, and 48% of those that take it a third time pass it then. (Assume that all students who do not pass the first or second time elect to take the test again.) What percent of the students take the test three times? (Round your answer to one decimal place.)
Mathematics
1 answer:
Alborosie3 years ago
6 0

Answer:

  14.4%

Step-by-step explanation:

Of the 39% who fail the first exam, 37% fail a second time. Those are the ones who take it 3 times:

  0.39 × 0.37 = 0.1443 ≈ 14.4%

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Answer:

The critical value is T = 4.604.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

Step-by-step explanation:

The first step to solve this question is finding the sample mean and sample standard deviation:

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Sample mean is sum of all values divided by the number of values. Thus:

\overline{x} = \frac{14+11+18+9+19}{5} = 14.2

The sample standard deviation is the square root of the division of the sum of the subtractions squared of each value and the mean, and the number of values. Thus:

s = \sqrt{\frac{(14-14.2)^2+(11-14.2)^2+(18-14.2)^2+(9-14.2)^2+(19-14.2)^2}{5}} = 3.868

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We have the standard deviation for the sample, and so we use the t-distribution to build the confidence interval.

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

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99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 4.604.

The critical value is T = 4.604.

The margin of error is:

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The lower end of the interval is the sample mean subtracted by M. So it is 14.2 - 7.964 = 6.236.

The upper end of the interval is the sample mean added to M. So it is 14.2 + 7.964 = 22.164.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

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