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Genrish500 [490]
2 years ago
12

Ms. Martino always gives a very easy 10-question quiz in the first week of her classes. Over the years, the number of questions

that students answer correctly on these quizzes has been strongly skewed to the left with a mean of 9 correct answers and a standard deviation of about 2.5 correct answers.
Suppose we took random samples of 4 students and calculated -x as the sample mean number of questions that the students answered correctly. We can assume that the students in each sample are independent.

What would be the shape of the sampling distribution of -x?
Mathematics
1 answer:
Natalka [10]2 years ago
4 0

Answer:

Skewed Left

Step-by-step explanation:

The shape of the sampling distribution of -x will match the shape of the parent population since the sample size is small (n=4<30, although it will be closer to normal than the parent population is.

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4

Step-by-step explanation:

100 divided by 25 equals 4.

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Solve 5x − 6y = −38 <br> 3x + 4y = 0
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What is the quotient of 5/6÷3/4?
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Simplify: √27 A. 3√3 B. 9√3 C. 3√9 D. 5√3
blagie [28]
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Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc
Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when M = 0.3, \sigma = 4.103. So

M = z*\frac{\sigma}{\sqrt{n}}

0.3 = 1.96*\frac{4.103}{\sqrt{n}}

0.3\sqrt{n} = 1.96*4.103

\sqrt{n} = \frac{1.96*4.103}{0.3}

(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

n = 718.57

Rouding up

We need a sample size of at least 719

6 0
3 years ago
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