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3 years ago

What is the mean absolute deviation of the data set?

Mathematics

1 answer:

Ronch [10]3 years ago

5 0

Answer:

the mean absolute deviation of the data set is 2.

Step-by-step explanation:

To calculate the mean absolute deviation of a data set, we need to follow the following steps:

Step 1: Calculate the mean.

Step 2: Calculate how far away each data point is from the mean (always keeping a possitive distance).

Step 3: Calculate the mean of the deviations.

Step 1: The mean of the data set is:

Mean = (12 + 10 + 10 + 8 + 6+ 7 + 7 + 12)/8 = 72/8 = 9

Step 2:

{12 -9, 10-9, 10-9, 9-8, 9-6, 9-7, 9-7, 12-9} = {3, 1, 1, 1, 3, 2, 2, 3}

Step 3: (3 + 1 + 1 + 1 + 3 + 2 + 2 + 3)/8 = 2

In conclusion, the mean absolute deviation of the data set is 2. Which corresponds to the option B.

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Ne4ueva [31]

Answer:

a) $n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer

b) For this case if we see the formula for the margin of error

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean

$\sigma=16$ represent the population standard deviation

n represent the sample size

Solution to the problem

Part a

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer

Part b

For this case if we see the formula for the margin of error

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

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