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Lubov Fominskaja [6]
3 years ago
12

If your score on your next statistics test is converted to a z​ score, which of these z scores would you​ prefer: minus−​2.00, m

inus−​1.00, ​0, 1.00,​ 2.00? why?
a. the z score of minus−2.00 is most preferable because it is 2.00 standard deviations below the mean and would correspond to the highest of the five different possible test scores.
b. the z score of 1.00 is most preferable because it is 1.00 standard deviation above the mean and would correspond to an above average test score.
c. the z score of 2.00 is most preferable because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible test scores.
d. the z score of minus−1.00 is most preferable because it is 1.00 standard deviation below the mean and would correspond to an above average test score.
e. the z score of 0 is most preferable because it corresponds to a test score equal to the mean.
Mathematics
1 answer:
Gemiola [76]3 years ago
3 0
The answer is z score of 2.00 is most desirable because it is 2.00 standard deviations beyond the mean and would agree to the highest of the five diverse likely test scores. In addition, the number of standard deviations that a given value x is above or below the mean is the z score or as called as standardized value. The negative z score agrees to an x value less than the mean while a positive z score agrees to an x value larger than the mean. The more negative the z score the more the x value is under the mean and the further positive the z score the more the x value is below the mean.
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Answer:

8 men.

Step-by-step explanation:

12÷3=4.

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4 years ago
A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25
GREYUIT [131]

Answer:

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

Step-by-step explanation:

The standard deviation is the square root of the variance. Since the variance is 25, the sample's standard deviation is 5.

We have the sample standard deviation, not the population, so we use the t-distribution to solve this question.

T 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

df = 15 - 1 = 14

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of 0.95(t_{95}). So we have T = 1.761

The margin of error is:

M = T*s = 1.761*5 = 8.805.

The upper end of the interval is the sample mean added to M. So it is 75 + 8.805 = 83.805.

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

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4 years ago
Write the phrase as an expression. 28 divided by 7
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Hope this helps

Have a great day/night

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