2. Understand key measures of central
tendency
• Mean
• Median
• Mode
Understand key measures of dispersion
• Normal Distribution
• Skew
• Standard Deviation
• Z Scores
3. We often want to know, what’s the typical,
more representative value of a variable
Examples:
Which gender is more represented in the
sample?
Which of our products is the most
popular
What is the average selling price?
What is the average initial salary?
4. Mean = the sum of all the members of the
list divided by the number of items in the
list
Median = the number separating the
higher half of a sample from the lower
half.
Mode = the most frequent value
5. A probability distribution that plots all of its values in a
symmetrical fashion and most of the results are situated
around the probability's mean
8. In addition to the most common value, we often
want to know how a sample is distributed
Jim’s order was $3. How common is that?
Tia ordered $35. How common is that?
Ed ordered $200. How common is that?
9. The most common measure of dispersion
1. Calculate the group mean ( )
(average order =$35)
2.Take everyone in the sample (Xi)
(Jim ordered $3 Tia ordered $35, Ed ordered $200, …)
3. Measure how much each one differs from the mean
(Xi - )
(Jim’s diff = -$32 Tia’s diff = $0, & Ed’s diff = $165)
4. Square all diff values & add them up
(1024+0+27225+……)
5. Divide that total by the sample size (N=310)
6.The result is the standard deviation.
10. The first SD covers the first 34.1% around the mean.
Two SDs above & below the mean covers 95% of the
distribution.
50th percentile16th percentile 84th percentile
11. Jim’s order was $3. He’s around -1 SD
Tia ordered $35. She’s an average customer
Ed ordered $200. $200-$35=$165
$165/$32 = 5.15 SD!
Ed’s extremely weird!
-1 Standard Deviation
$34.72 (mean)-$32 (SD) = $2.72
Mean $34.72 = tip of bell curve
5.15 Standard Deviation
$34.72 (mean)+ 5.15 * $32 (SD) = $200
12. Jim’s order was $3. Jim’s z score is -1
Tia ordered $35. Tia’s z score is 0
Ed ordered $200. $200-$35=$165
$165/$32 = 5.15 SD!
Ed’s z score is 5.15
-1 Standard Deviation
$34.72 (mean)-$32 (SD) = $2.72
Mean $34.72 = tip of bell curve
5.15 Standard Deviation
$34.72 (mean)+ 5.15 * $32 (SD) = $200
Editor's Notes
This lecture presentation complements Khan’s tutorials.
In this lecture we will discuss the different methods to measure central tendency and dispersion in a statistical sample.
Central tendency is just a technical way of saying, what’s typical of this sample? For example, out of all Carlow students, which gender is the more typical one? Male or female? Out of all the products listed on Amazon, which is the best seller? And out of all the eBay listings of “Tickle Me Elmo,” which price is the most common one?
These three different measures are discussed in detail by Khan Academy. Here are some brief summaries. We will discuss normal distribution. One key idea is this: If the sample is normally distributed, meaning it looks like a symmetrical bell curve, then mean, median and mode will be the same number. However, if the sample is skewed either to the left or to the right, then these three numbers would take on different values.
Concepts like “mean” and “standard deviation” are really based on the theory of normal curve. Note it’s a theory, a conceptualization of how data should be distributed in an ideal world. In reality, often times distributions are not perfectly normal. Next slide is an example. Note that the “mean” = the 50th percentile.
Look at this distribution of salary data. It’s heavy on the left side, with a long skinny tail on the right. Definitely not symmetrical.
When we impose the normal curve on top of the salary distribution, we see that the normal curve only captures the right tail well. For the left tail, the normal curve doesn’t describe the actual distribution very well. This is because the salary data is positively skewed. In skewed data, “mode” and “median” describe the central tendency better than the “mean”.
In addition to central tendency, we also need a way to describe how spread out the distribution is, and how weird a case is (relative to the mean). When a case is very close to the mean, we have an average joe. When a case is far off from the mean on the tip of a long tail, we have a weirdo! In real life, we often discuss dispersion without realizing it. For example: In which percentile is my child’s height? How many people in this class will get an A? Is the customer’s credit score above or below average? By how much? Is a donation of $30,000 pretty common or very rare? How rare is it? This slide illustrates the distribution of total purchase after a customer clicks on a link. Look at the data, the mean, the distribution, and reflect on the following questions: How likely would an average customer spend $200 per order? Very unlikely – it’s at the end of the curve – in a tail. How about $35? Much more likely – it’s the average order. In what percentile is a $67 order? The 84th - we know because it’s one standard deviation (34%) above the mean (50%). The next slide explains what a standard deviation is.
Standard deviation is a standardized measure of dispersion. It tells you whether the distribution is short and fat (with a big standard distribution) or tall and skinny (with a small standard distribution). The calculation is explained well by Khan (see Khan’s Academy video clips linked in this session). The basic idea to take away is: The standard deviation tells you, on average, how far away the data points are from the mean. For example, let’s say that the Steelers have an average score of 25 per game, and the standard deviation is 1. Let’s also say that the Greenbay Packers have an average score of 25 per game, and a standard deviation of 7. In this example, both teams are comparable in terms of average scores, but the Steelers have a much smaller standard deviation. This means the Steelers’ performance is pretty consistent over time, their scores may be above or below 25, but only by 1-2 points on average. If you plot their scores on a chart, you would see that most of them pack around 25, with a nice narrow distribution that peaks around 25. In contrast, the Packers may average around 25, but their performance varies widely from game to game. One day they may score 18 (25-7) and the next day they may score 32 (25+7) If you plot their widely varied scores on a chart, you would get a short and fat distribution. (Go Steelers Go!)
What are practical ways to use the standard deviation? With a normal distribution, the mean divides it up evenly in the middle. The portion below the mean covers 50% of the population, whereas the portion above the mean also covers 50% of the population. The first standard deviation away from the mean covers 34% of the distribution. In other words, 1 standard deviation above the mean = 50% + 34% = 84% = 84th percentile Let’s say that the average weight for a one year old is 25 lbs, with a standard deviation of 2 lbs. Connor is 23 lbs. That’s 1 standard deviation below the mean. In other words he is 50%-34% or in the16 th percentile of the population Nardia is 27 lbs. That’s 1 standard deviation above the mean. In other words she is 50%+34% or in the 84 th percentile of the population The entire distribution is covered by roughly 6 standard deviations – 3 above the mean and 3 below the mean Hence the name of the quality management program “ Six Sigma ”
More examples: Given a mean and a standard deviation score, you have a pretty good idea of what the distribution is like – is it fat and short, or tall and skinny? We can then map out individual scores on the distribution and tell the average joes from the weirdos!
The Z score is the number of standard deviations from the mean. With our previous example, Connor would have a Z score of negative 1 (that is 1 standard deviation below the mean), while Nardia has a Z score of 1 (that is 1 standard deviation above the mean). The average joes would have close to zero z scores (e.g., 0.0006, -.0029) Whereas the weirdos have extremely large or small z scores (e.g., 3.07, -2.99) Again - The z score is the number of standard deviations that a data point is away from the mean. Let's say that the average weight for all American women is 150 lbs, and the standard deviation is 20 lbs. If your weight is 130, then your z score is -1, because you're exactly 1 standard deviation below the mean. If Peggy's weight is 170, then her z score is 1, because she is exactly 1 standard deviation above the mean.
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