Determine the mean, standard deviation, and five number summary of each data set round to the nearest hundredth, if necessary
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I'll do problem 3 to get you started.

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Part 1) Finding the mean:

The mean is found by adding up all the numbers and dividing by n, where n is the number of items in the set.

We sum the numbers to:

5+8+5+9+6+15+9+3+8+7+10+12 = 97

Then we divide over n = 12

mean = sum/n = 97/12 = 8.083333 approximately

This rounds to 8.08

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Part 2) Finding the standard deviation:

I'm assuming your teacher is talking about the population standard deviation, as opposed to the sample standard deviation.

First subtract each value from the mean. It helps to make a table as shown in the attachment below.

Column A is the original data set. Column B is when we subtract each data item x from the mean M.

Column C is where we square each item in column B.

Adding up everything in column C leads to 118.916666666667 which is approximate. This value is highlighted in the table diagram. Divide this over n = 12 to get 9.90972222222222 which is also approximate

The last step is to apply the square root to that and we get 3.1479711279

This rounds to 3.15

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Part 3) Finding the min:

From here on out, we have fairly simple calculations and don't need to involve a spreadsheet or any complicated formulas.

The min value is the smallest value of the set. So that would be 3

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Part 4) Finding the value of Q1, first quartile:

Sort the items in ascending order, ie from smallest to largest.

You should get this set: {3, 5, 5, 6, 7, 8, 8, 9, 9, 10, 12, 15}

There are n = 12 items here, so the median is between slot 12/2 = 6 and slot 7. The values in slot 6 and 7 are the two copies of '8'. That must mean the median is (8+8)/2 = 8.

Now break up the set into two equal halves

L = {3, 5, 5, 6, 7, 8} = lower half of six items

U = {8, 9, 9, 10, 12, 15} = upper half of six items

The median of set L is exactly the value of Q1, so that would be (5+6)/2 = 11/2 = 5.5; note how this is exactly between the two middle most values of 5 and 6.

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Part 5) Finding the median:

We found that earlier in part 4. The median is 8.

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Part 6) Finding the value of Q3, third quartile:

Focus on set U = {8, 9, 9, 10, 12, 15}, which was the upper set of values, ie values larger or equal to the median.

The middle two values here are the '9' and '10'. The median is at the midpoint of these values: (9+10)/2 = 19/2 = 9.5

We have Q3 = 9.5

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Part 7) Finding the max:

The max is the largest item in the original set

Refer to the sorted set {3, 5, 5, 6, 7, 8, 8, 9, 9, 10, 12, 15} and we see that 15 is the max.

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Part 8) Finding the IQR (interquartile range):

Subtract the values of Q3 and Q1 to get the IQR

IQR = Q3 - Q1

IQR = 9.5 - 5.5

IQR = 4

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Part 9) Finding the range:

The range is a similar idea as the IQR; however, we'll subtract the max and min this time.

Range = max - min = 15 - 3 = 12

The range is 12.

The range, IQR and standard deviation all help us determine the spread of the data. The larger the value gets, the more spread out the data is.

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Summary:

Side note: The phrase "five number summary" refers to the five values of: min, Q1, median, Q3, max. All of which are needed to form a box-and-whisker plot.

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