unit 4: describing data

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Unit 4: Describing Data. Central Tendency S.ID.2: Use statistics appropriate to the shape of the data distribution to compare center and spread of two or more different data sets. Essential Question: How do I find the mean, median, mode, and range in a numerical set of data?. Vocabulary: - PowerPoint PPT Presentation

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Unit 4: Describing Data

Central Tendency

S.ID.2: Use statistics appropriate to the shape of the data distribution to compare center and spread of two or more different data sets.

Essential Question:

How do I find the mean, median, mode, and range in a numerical set of data?

Vocabulary:

Mean: the average value from a numerical set of data. The symbol is x, which is read as “x-bar”.Median: the middle number from a numerical set of data when written in order. If the data set has an even number of values, the median is the mean of the two middle values. Also known as the “quartile”.Mode: The mode is the value that occurs most frequently in a data set. There can be one mode, or many. There can also be no mode in a data set.Range: In a numerical set of data, the range is the difference between the highest value and the lowest value.Measure of Dispersion: Describes the dispersion, or spread, of data.Deviation from the Mean: The difference of a data value and the mean of the data set.

TEST SCORES: The test scores received by students on a history exam are listed below. Find the mean, median, mode, and range:65, 68, 71, 77, 81, 82, 86, 88, 93, 93, 95, and 97MEAN = MEDIAN = MODE = RANGE =

Example 1:

Which measure(s) best represent the data?65, 68, 71, 77, 81, 82, 86, 88, 93, 93, 95, and 97MEAN = 83MEDIAN = 84MODE = 93RANGE = 32

Mean and Median BEST represent the data given.

Class work:

Red Math 1 books that are under your desk.

Pg. 365, #’s 1-4 (find the mean, median, mode, and range)

Lastly, do # 15 a, b, and c

1. Mean = 5 Median = 5 Modes = 1, 5 Range = 102. Mean = 19, Med = 22, Mode = 25, Range = 183. Mean = 16, Med = 13.5, Modes = 8, 28 Range = 224. Mean = 2.8, Med = 2.8, Mode = none Range=3.115a. 2415b. Mean = 41.25, Med = 41.5, Mode = none15c. Both mean and median represent the data well – they are both close to all data points.

Answers:

Homework:

Worksheet

HW Answers:

1. Mean: 58 2. Mean: 56Median: 59Median: 55Mode: noneMode: noneRange: 85

Range: 78

3. Mean: 45 4. Mean: 56Median: 36Median: 50Mode: noneMode: noneRange: 85

Range: 755. Mean: 61 6. Mean: 66Median: 65Median: 79.5Mode: noneMode: 80, 81Range: 86

Range: 827. Mean: 45 8. Mean: 75Median: 41Median: 80Mode: noneMode: noneRange: 87

Range: 819. Mean: 48 10. Mean: 55Median: 43.5Median: 53Mode: noneMode: 96Range: 85

Range: 83

Essential Question:

How do I find the mean absolute deviation in a numerical set of data?

Standard: S.ID.2

Finding the mean absolute deviation:Find the mean of the data setSubtract the mean from each individual piece of data. Take the absolute value of your answer (MAKE IT POSITIVE!).Add all of those numbers together and divide by the total number of data entries. Round to the nearest hundredth.This value is the mean absolute deviation.

Example 2:

Find the mean absolute deviation of the data set: 67, 69, 69, 71, 74, 76MEAN =

Class work:

Textbook pg. 365, #’s 9-14 (just find the mean absolute deviation)

9. 1.6810. 2.8811. 4.6712. 2.2513. 2.4414. 0.6

HW: Worksheet

QuartilesEssential Questions: How do I find the lower (Q1) and upper quartile (Q3) and the interquartile range?How do I compare summary statistics?

Quartile: the median of an ordered data set (Q2)

Upper quartile: the median of the upper half of an ordered data set (Q3)

Lower quartile: the median of the lower half of an ordered data set (Q1)

Interquartile Range: the difference of the upper quartile and lower quartile of an ordered data set

Vocabulary:

Example 1:

The data sets below give the number of home runs by each player on the Bears and the Wildcats during a season of the Oakmont Baseball League. Compare the data using the mean, median, range and interquartile range.

Bears: 28, 25, 21,19, 18, 14, 10, 8, 7 , 5, 3, 2Wildcats: 20, 19, 18, 16, 15, 15, 12, 11, 9, 8, 6, 5, 4

Solution:

Bears:Wildcats:Mean: 13.33 Mean: 12.15Median: 12Median: 12Range: 26Range: 16Lower Quartile: 6 Lower Quartile: 7Upper Quartile: 20 Upper Quartile: 17Interquartile Range: 14 Interquartile Range: 10

The Bears’ mean is greater than the Wildcats’ mean so they averaged more homeruns per player than the Wildcats.The Wildcats’ range is less than the Bears’, so their data is less spread out than the Bears’ data.The WC interquartile range is less than the Bears’ interquartile range, so the WC middle 50% of the data showed less variation than the middle 50% of the Bears’ data.

Solution:

Average Mean = 10.5 + 11.6 + 13.4 =

311.83Average median = 12.17Average Range: 25.33Average Interquartile Range: 15.33

What does the data mean??

Class work:

Red Mathematics 1 Textbook: pg. 371, #’s 1-4, 6, 8-9

Answers:

1. Mean= 8 Med = 8 R = 9 LQ = 5.5 UQ = 11 IQR= 5.52. Mean= 60.3 Med = 61 R = 55 LQ = 49 UQ = 71 IQR= 223. Mean= 4.7 Med = 4.9 R = 5.2 LQ = 3.1 UQ = 6.3 IQR= 3.24. Mean= 115 Med = 110 R = 119 LQ = 77 UQ = 148.5 IQR= 71.56. Answers may vary8. Avg. Mean = 86.6Avg. Median = 87.4Avg. Range = 34.6Avg. LQ = 80.2Avg. UQ = 92.4Avg. IQR = 12.2Avg. Mean is greater than pop. The avg. median, range, and IRQ are less.9. Avg. Mean = increases to 87.3Avg. Median = increases to 88.75 and is now greater than the population median.Avg. Range = decreasesAvg. IQR = decreases

Central Tendency Lab Instructions:

Form groups containing 5-6 people per group (move desks accordingly).

Measure the height of each group member in inches with a tape measure or ruler (I will provide those).

Record those heights (including yours) in the table provided on problem # 3.

Complete the worksheet front and back.

Each student will be turning in a completed worksheet.

Lab sheet answers:

Essential Questions:

How do I create a dot plot from a numerical data set?

How do I create a box and whisker plot from a numerical data set?

Discrete data - data that has only a finite number of values.Dot Plot - a graph that shows how discrete data are distributed using a number line.Data distribution - the way in which the data is spread out or clustered together.

symmetric - the left and right halves of the graph are nearly mirror images of each other.

skewed right - the peak of the data is to the left side of the graph.

skewed left - the peak of the data is to the right side of the graph.

Twenty high school students were randomly selected from a very large high school. They were asked to keep a record for a week of the number of hours they slept each night. These seven values were averaged to obtain an average night of sleep for each. The results are as follows: 9, 8, 8, 7.5, 6, 6, 4, 5.5, 7, 8, 5, 7.5, 6.5, 10, 8.5, 6.5, 5, 5.5, 7, and 7.5 hours. Create a dotplot of these data and discuss what the display implies about the data.

The ages of the Oscar winners for best actor and best actress (at the time they won the award) from 1996 to 2004 are as follows: 45, 39, 59, 33, 45, 25, 42, 24, 35, 32, 46, 32, 28, 34, 42, 27, 36, and 30. Create a doplot of these data and discuss what the display implies about the data.

box-and-whiskers plot - displays the data distribution on a five number summary.The five number summary consists of:

minimum valueQ1 (the first quartile)medianQ3 (the third quartile)maximum value

Construct a box and whiskers plot for the data set: {5, 2, 16, 9, 13, 7, 10}

Construct a box and whiskers plot for the data set:5 9 13 9 9 14 7 3 6 8 8 6 9 8 5

Assignment:

Pg. 453-458, #'s: ALL

OMIT #'S 7 AND 8 ON PG. 456!!!!!

Answers:

Find the mean, median, mode, range, IQR, make a dot plot and a box and whisker plot for the following data set:

105, 101, 102, 99, 110, 90, 100, 100, 108, 120, 32

Mean: 97Median: 101Mode: 100Range: 88IQR: 9

32, 90, 99, 100, 100, 101, 102, 105, 108, 110, 120Five # Summary:Minimum: 32Q1: 99Median: 101Q3: 108Maximum: 120

Outliers

EQ: How do I find outliers and what effects do they have on the context of a data set?

Standard: S.ID.3 - Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).

Outlier: a data value that is significantly greater or lesser than other data values in a data set.

You are determining a lower and upper limit for the data. Any value outside of these limits is an outlier.

The lower limit is called the "lower fence".

The upper limit is called the "upper fence".

Calculating the Fences:

lower fence: Q1 - (IQR x 1.5)

upper fence: Q3 + (IQR x 1.5)

Example 1: Make a box and whisker plot (find the five number summary) then calculate the upper and lower fences to determine if there are any outliers in the data set.

2, 5, 6, 6, 7, 9, 10, 11, 12, 12, 14, 28, 30

Five # SummaryMinimum Value:Q1:Median (Q2): Q3: Maximum Value:

Box and Whisker Plot:

Lower Fence: Q1 - (IQR x 1.5)

Upper Fence: Q3 + (IQR x 1.5)

Are there any outliers?

The idea is if any outliers are discovered, that you would remove them and recalculate the values of central tendency to decrease the spread of your data.

Example 2: Same directions as example 1.

10, 13, 17, 20, 22, 24, 24, 27, 28, 29, 35

Five # SummaryMinimum Value:Q1:Median (Q2): Q3: Maximum Value:

Box and Whisker Plot:

Lower Fence: Q1 - (IQR x 1.5)

Upper Fence: Q3 + (IQR x 1.5)

Are there any outliers?

Example 3: Same directions as example 1.

0, 7, 17, 17, 18, 24, 24, 24, 25, 27, 45

Five # SummaryMinimum Value:Q1:Median (Q2): Q3: Maximum Value:

Box and Whisker Plot:

Lower Fence: Q1 - (IQR x 1.5)

Upper Fence: Q3 + (IQR x 1.5)

Are there any outliers?

Assignment:Pg. 467, #'s 1-4 AND pg. 468-473, #'s 1-12

HW answers:

1. b 2. d 3. a 4. c

2. IQR = 10 LF: 1 UF: 41 0 is an outlier

3. IQR = 12 LF: 10 UF: 589 and 59 are outliers

4. IQR = 11 LF: 15.5 UF: 59.5no outliers

5. IQR = 6.5 LF: 8.75 UF: 34.758 is an outlier

6. IQR = 22.5 LF: 21.25 UF: 111.2515, 20, and 115 are outliers

8. IQR = 4 LF: 10 UF: 26no outliers

9. IQR = 15 LF: 22.5 UF: 82.5At least 1 outlier on the lower side and at least 1 outlier on the upper side.

10. IQR = 5 LF: -.5 UF: 19.5At least 1 outlier greater than the upper fence.

11. IQR = 200 LF: 50 UF: 850At least 1 outlier less than the lower fence.

12. IQR = 6 LF: -5 UF: 19No outliers.

Class work assignment:

Pg. 103-105 ALL (thin books)

**problem 1 d. wants you to reconstruct the box and whisker plot by removing the outliers you found and then recalculating your five number summary to create a new box and whisker plot.

Answers:

EQ: How do I use and analyze histograms?

Histogram: a graphical way to display quantitative data using vertical bars.

Bin: intervals of data along the horizontal axis of a histogram

Frequency: the height of each bar (vertical axis) which is the number of data values included in each bin

Bin Intervals: bin intervals can be written as compound inequalities.

For example: 0, 5, 10, 15, 20, 25 across x-axis

Data points: 2, 4, 5, 6, 15, 23, 10, 11, 14, 12, 24

Get a Student Text Volume 2 book...

Turn to page 463 and look at problem #3.

We are going to walk through some examples together.

Please draw the histogram at the bottom of pg. 463 in your notes.

2.3 1.8 10.2 7.0 29.4 12.4 15.3 21.1 5.2 12.1 41.9 22.1 9.3 15.2 18.4 21.6 12.0 10.8 31.5 15.9 6.2 30.0 23.6 17.6 12.3 15.2 19.4 8.1 32.5 29.1 21.4 12.0 9.3 16.3 22.7

Histogram Problem A number of students were asked to calculate their average commute time one-way to school. Their averages (in minutes) are listed in the table below.

Create a histogram with bin intervals of 5 minutes.

How many modes are their? List all modal groups.

Describe the shape of the histogram.

Make some conclusions about driving times based on your graph.

Class work / Homework:

pg. 97-100 (thin book), #'s 1-2 ALL

Answers:

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