Lecture 5
Farmingdale State College
They are shaped like a bell (“bell curve”).
They are symmetric.
They are unimodal.
The mean = median = mode.
x
).98
on the exam. The average grade on the exam was 97
with a standard deviation of 1
.
x
).98
on the exam. The average grade on the exam was 97
with a standard deviation of 1
.
68.3% of the data will fall within 1 SD of the mean.
95.4% of the data will fall within 2 SD of the mean.
99.7% of the data will fall within 3 SD of the mean.
“Within one standard deviation” means +1 as well as -1 standard deviation.
Students’ ratings of the Netflix Original Dark (range = 1- 10) form a normal distribution with m = 6 and s = 1.
What percentage of Students rate it a 7 or higher?
What percentage of Students rate it at least a 4?
What percentage of Students rate it an 8 or lower?
Students’ ratings of the Netflix Original Dark form a normal distribution with m = 6 and s = 1.
What percentage of Students rate it a 7 or higher?
15.8% (13.6 [1SD] + 2.1[2SD] + .1[3SD])
Students’ ratings of the Netflix Original Dark form a normal distribution with m = 6 and s = 1.
What percentage of Students rate it at least a 4?
2.2%
Students’ ratings of the Netflix Original Dark form a normal distribution with m = 6 and s = 1.
What percentage of Students rate it an 8 or lower?
97.6%
A z-score tells you, in standard deviation units how far the x-value is from the mean.
Z-scores are better than using raw SD
When the SD is a decimal, it is hard to find the exact point under the standard normal curve.
Z-scores re-express the original data points (the x
’s) in a way that intuitively lets us know:
How close the x
is to the mean (how much this particular participant is like the average person in the sample)
Where it falls in the dispersion of the distribution (how different this particular participant is from the majority of people in the sample)
Formula
\[ z = \large{\frac{\color{orange}{x}-\color{red}{\bar{x}}}{\color{pink}{s}}} \]
Subtract each x value from the mean.
Divide by the standard deviation
z = \(\frac{x - \mu}{\sigma}\)
Mean (\(\mu\)) = 50
Standard deviation (\(\sigma\)) = 10
Raw score (x) = 65
Find the z-score.
\(\frac{65-50}{10} = 1.5\)
Mean (\(\mu\)) = 100
Standard deviation (\(\sigma\)) = 15
Raw score (x) = 85
Find the z-score.
\(\frac{85-100}{15} = -1\)
Mean (\(\mu\)) = 70
Standard deviation (\(\sigma\)) = 8
Raw score (x) = 66
Find the z-score.
$ = .5
Mean (\(\mu\)) = 500
Standard deviation (\(\sigma\)) = 120
Raw score (x) = 740
Find the z-score.
\(\frac{740-500}{120} = 2\)
Mean (\(\mu\)) = 40
Standard deviation (\(\sigma\)) = 6
z-score = +2.5
Find the raw score (x).
\(2.5 = \frac{x-40}{6}\)
\(x = z\times s = 2.5\times6=15\)
Centering the X values on the mean: When we center the mean (mean-centering), we set the mean to 0.
Dividing by the standard deviation
When we divide by the SD, the space from the mean is expressed in standard deviations.
\(x\) | \(\bar{x}\) | \(x-\bar{x}\) | \((x-\bar{x})^2\) |
---|---|---|---|
10 | 20 | −10 | 100 |
10 | 20 | −10 | 100 |
20 | 20 | 0 | 0 |
30 | 20 | 10 | 100 |
30 | 20 | 10 | 100 |
Average
\[\color{red}{\bar{x}} = \frac{\sum{x}}{n}\]
Standard Deviation
\[s^2 = \sqrt{\frac{\sum(x-\color{red}{\bar{x}})^2}{n-1}} = \frac{400}{4} = \sqrt{100} = s\]
Z-Score
\[z =\frac{x-\color{red}{\bar{x}}}{s} = \frac{20-10}{s}=\frac{-10}{s}\]
\(x\) | \(x-\bar{x}\) | \((x-\bar{x})^2\) | \(z=\frac{(x-\bar{x})^2}{s}\) |
---|---|---|---|
10 | −10 | 100 | −1 |
10 | −10 | 100 | −1 |
20 | 0 | 0 | 0 |
30 | 10 | 100 | 1 |
30 | 10 | 100 | 1 |
We made the mean = 0: When you mean-center a distribution, you shift it along the number line.
We made the SD = 1: When you divide a distribution by the SD, you shrink the distribution down.
The shape of the distribution remains the same.
Shift distribution along the number line.
Shrink distribution down.
The shape of the distribution remains the same.
A z-score tells me where my score falls in SD units.
I can then look at this standard normal curve, and estimate what percentage of people did better or worse than me.
The mean score for Exam 1 was a 92 with a standard deviation of 3.
Esmeralda scored an 86.
What percent of the class scored better than Esmeralda?
\(z=\frac{x-\bar{x}}{s} = \frac{86-92}{3} = -2\)
98% of the class did better than Esmeralda.
The mean score for Exam 1 was a 92 with a standard deviation of 3.
Jonas scored a 95.
What percent of the class scored better than Jonas?
\(z = \frac{95-92}{3} = 1\)
16% scored higher than Jonas.
NULL
Standard Normal Distrubution | ⬡⬢⬡⬢⬡