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2 years ago

The scores listed below are the results from students taking a dexterity test. What is the 75th percentile of these scores?

2 answers:

8 0

Step 1 find the median of thos data set

the middle number is 2

then find the median of the right side of the data that is left that is the 75th percentile
which is 2.25

meriva2 years ago

6 0

The 75th percentile is also called the 3rd quartile.

First find the median, since this is an odd number of elements it is simply the middle term (if there are an even number of terms it will be the average of the two middle terms) which in this case is 2.

Now find the median term between 2 and the last term, in this there are an even number of terms so it will be the average of the two middle terms 2 and 2.5, which is:
2.25

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3 years ago

Read 2 more answers

"A laboratory tested 73 chicken eggs and found that the mean amount of cholesterol was 230 milligrams with LaTeX: \sigmaσ = 17.4

Scrat [10]

Answer:

The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.

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 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.

$M = 1.96*\frac{17.4}{\sqrt{73}} = 3.99$

The lower end of the interval is the sample mean subtracted by M. So it is 230 - 3.99 = 226.01

The upper end of the interval is the sample mean added to M. So it is 230 + 3.99 = 233.99.

The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.

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3 years ago