1. Chapter 6: The Standard Deviation as a Ruler and the Normal Model
By Marisa Suzuki

2. Standard Deviation as a Ruler
Standard deviations can be used to compare very different-looking data
Standard deviation tells us how the whole collection of values varies
The more variability in data, the higher the standard deviation will be

3. Standardizing with z-scores
Results can be standardized, resulting in values denoted with the letter z, but are usually called z-scores
Z-scores measure the distance of each data value from the mean in standard deviations
Can be found using the equation:

4. Standardizing with z-scores cont.
The mean of all of the data is subtracted from the sample value y, which is then divided by the standard deviation.
A negative z-score tells us that the data is below the mean, while a positive z-score tells us that the data is above the mean

5. Combining Two Variables
To find the standard deviation of the sum or difference, you must ADD the variances and then take the SQUARE ROOT

6. Normal Models
Distributions that are unimodal and roughly symmetric can be represented by a normal model (bell-shaped curve)
N(μ,σ) denotes a normal model
µ = (mew) mean of the model
σ = (lower case sigma) standard deviation in the model
The standard normal model is a model with a mean of 0 and SD of 1, denoted as N(0,1)

7. Normal Models cont.
Normal models give us an idea of how extreme a value is by telling us how likely it is to find one that far from the mean
The Rule
68% of the values fall within 1 standard deviation of the mean
95% of the values fall within 2 standard deviations of the mean
99.7% of the values fall within 3 standard deviations of the mean

8. The Bell-Shaped Curve

9. Question #35
Based on the normal model for weights of Angus steers N(1152,85), what are the cutoff values for:
The highest 10% of the weights?
The lowest 20% of the weights?
The middle 40% of the weights?

10. Question #35 (part a) 2ND  DISTR  invNorm (0.9, 0, 1)  1.282
The highest 10% of the weights?
2ND  DISTR  invNorm (0.9, 0, 1)  1.282
z=(y-µ)/σ
1.282=(y-1152)/84
y= lbs

11. Question #35 (part b) 2ND  DISTR  invNorm (0.2, 0, 1)  -0.842
The lowest 20% of the weights?
2ND  DISTR  invNorm (0.2, 0, 1) 
z=(y-µ)/σ
-0.842=(y-1152)/84
y= lbs

12. Question #35 (part c) 2ND  DISTR  invNorm (0.3, 0, 1)  -0.524
The middle 40% of the weights?
2ND  DISTR  invNorm (0.3, 0, 1) 
2ND  DISTR  invNorm (0.7, 0, 1)  0.524
z=(y-µ)/σ
-0.524=(y-1152)/84
y= lbs
0.524=(y-1152)/84
y= lbs

13. Question #37 Consider the Angus steer model N(1152,84) again:
What weight represents the 40th percentile?
What weight represents the 99th percentile?
What’s the IQR of the weights of these Angus steers?

14. Question #37 (part a) z=(y-µ)/σ -0.253=(y-1152)/84 y=1130.7 lbs
What weight represents the 40th percentile?
2ND  DISTR  invNorm (0.4, 0, 1) 
z=(y-µ)/σ
-0.253=(y-1152)/84
y= lbs

15. Question #37 (part b) z=(y-µ)/σ 2.326=(y-1152)/84 y=1347.4 lbs
What weight represents the 99th percentile?
2ND  DISTR  invNorm (0.99, 0, 1)  2.326
z=(y-µ)/σ
2.326=(y-1152)/84
y= lbs

16. Question #37 (part c)
What’s the IQR of the weights of these Angus steers?
2ND  DISTR  invNorm (0.25 0, 1) 
2ND  DISTR  invNorm (0.75, 0, 1)  0.674
Q1 =
Q3 =
IQR = (Q3 – Q1) = ( – ) = lbs