Download Chapter 5 The Normal Distribution (Pt. 2)

Chapter 5
The Normal Distribution (Pt. 2)
5.1
Finding Normal Percentiles
Recall that the Nth percentile of a distribution is the value that marks off the
bottom N% of the distribution. For review, remember that Q1 is the 25th percentile, the median is the 50th percentile, and Q3 is the 75th percentile. A normal
percentile is simply a percentile of a normally-distributed variable.
In the last lecture, we learned about the empirical rule’s estimates for the normal
distribution’s central regions. Let us extend that rule to make some estimates
on normal percentiles. Let X ∼ N (µ, σ). Then, by application of the empirical
rule:
1. µ − 2σ is approximately the 2.5th percentile.
2. µ − 1σ is approximately the 16th percentile.
3. µ is approximately the 50th percentile (i.e. the median).
4. µ + 1σ is approximately the 84th percentile.
5. µ + 2σ is approximately the 97.5th percentile.
Refer to the illustrations in the last lecture if you are puzzled as to how we got
these numbers.
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
Note that if x marks the Nth percentile, then it follows straightforwardly that x
also marks off the top (100 − N)% of the distribution.
The table of standard normal percentiles or, more colloquially, the z-table, percentile information on the standard normal distribution. Thus, the percentiles in
this case are actually z-scores. The table ranges from −3.90 ≤ z ≤ 3.90 in increments of 0.01 and the percentiles are given to the fourth decimal place.
Given a z-score—say z = 1.15—you can find the percentile to which it corresponds by first looking down the left margin and locating 1.1. You then move
along the row next to 1.1 until you get to the entry in the column headed by 0.05.
This entry is the desired percentile—which, in this case, is 0.8749.
z
0.00
..
0.0
.
..
..
.
.
1.1 · · ·
..
..
.
.
..
3.9
.
···
..
.
..
.
···
..
.
..
.
···
..
.
..
.
0.8749 · · ·
..
..
.
.
..
..
.
.
0.05
..
.
..
.
0.09
..
.
..
.
···
..
.
..
.
Figure 5.1: How to look up a z-score’s percentile
At this point, we must introduce an important piece of notation. Since 87.49% of
the standard normal curve lies to the left of z = 1.15, it follows that if we were
to randomly choose some z-score, 87.49% of the time it would be less than 1.15.
That is, the probability statement
P( Z < 1.15) = 0.8749
holds true. In general, for a particular z-score z,
P( Z < z) = the z-table’s entry for z
(It is highly recommended that you get a handle on this notation ASAP). Also
note that, for technical reasons, P( Z < z) = P( Z ≤ z). That is, including z does
not make a difference.
Now, because any normal distribution (e.g. X ∼ N (µ, σ)) can be converted to the
standard normal distribution (e.g. Z ∼ N (0, 1)), we can take any value x from
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
X, convert it to a z-score, and then use the z-table to find out what percentile
x corresponds to in its original distribution. Therein lies the usefulness of the
standard normal distribution.
Example (Standard Normal Percentiles)
For the following, we use the model Z ∼ N (0, 1). For each problem, you should
first draw a bell curve and shade in the area of interest.
Determine the following:
1. The percentage of the standard normal curve below z = 2.24.
2. The percentage of the standard normal curve above z = 0.07.
3. P( Z < −1.02)
4. P( Z ≤ −1.02)
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
Example (Non-Standard Normal Percentiles)
For the following, we use the distribution X ∼ N (1152 pounds, 84 pounds) describing the weights of young Angus steers in Problem 5.19. As before, draw a
picture of the model and shade in the area of interest. Also, note that the z-table
only gives us z-values up to the second decimal place, so round accordingly.
Determine the following:
1. The percentage of Angus steers weighing less than 1250 pounds.
2. The probability of randomly choosing an Angus steer weighing less than
1000 pounds.
3. If a z-score is a measure of unusualness, then which would be more unusual: a steer weighing 1250 pounds or one weighing 1000 pounds?
4. If a particular Angus steer had a z-score of 1.35, how much did it weigh?
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
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5.2
Using Normal Percentiles to Find More General
Areas of the Normal Curve
The standard normal percentiles in the z-table give the proportion of the normal
curve less than, or to the left of, a given z-value. However, we might be interested in
the areas of more general areas of the standard normal curve, like the proportion
greater than a given z-score, or the region between two z-scores.
Here are the three basic rules for finding Normal curve areas using only percentiles. We name them Rule L(ess Than), Rule G(reater Than), and Rule
B(etween):
Rule L: Let z be a z-value. Then the proportion of the standard normal curve
less than z is simply its entry in the z-table. That is:
0.2
0.1
0.0
dnorm(x)
0.3
0.4
P( Z < z) = Entry in the z-table
-3
-2
-1
0
x
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
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Rule G: Let z be as before. The proportion of the standard normal curve greater
than z is 1 minus z’s entry in the table. That is:
0.2
0.0
0.1
dnorm(x)
0.3
0.4
P( Z > z) = 1 − Entry in the z-table = 1 − P( Z < z)
-3
-2
-1
0
1
2
3
1
2
3
0.2
0.1
0.0
dnorm(x)
0.3
0.4
x
-3
-2
-1
0
x
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Rule B: Let z1 < z2 be two z-scores. The proportion of the standard normal
curve between z1 and z2 is the difference between the areas left of z2 and z1 .
That is:
0.4
0.3
0.2
dnorm(x)
0.0
0.1
0.2
0.1
0.0
-1
0
1
2
3
-3
-2
-1
0
x
0.2
0.3
0.4
x
dnorm(x)
-2
0.1
-3
0.0
dnorm(x)
0.3
0.4
P(z1 < Z < z2 ) = Entry for z2 − Entry for z1 = P( Z < z2 ) − P( Z < z1 )
-3
-2
-1
0
x
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1
2
3
1
2
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
Example (Areas of the Standard Normal Curve)
For the following, use the standard normal model (Z ∼ N (0, 1)). As before, draw
a picture of the bell curve, shading in the area of interest.
Using the 3 rules, determine the following:
1. P( Z > 0.65)
2. P(| Z | < 1.01) = P(−1.01 < Z < 1.01)
3. P(−0.25 < Z < 2.25)
4. P( Z = 1.12)
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Example (Areas of Other Normal Distributions)
For the following, we use the previous example of X ∼ N (1152 pounds, 84 pounds)
describing the weights of Angus steers.
Determine the following:
1. The percentage of steers weighing more than 1200 pounds.
2. The percentage of steers weighing between 1100 and 1300 pounds.
5.3
Solving for Missing Parameters
Recall the direct calculation of z:
z=
x−µ
σ
This equation can be re-expressed in three other ways by isolating the other
variables on the left hand side:
x = zσ + µ
x−µ
σ=
z
µ = x − zσ
In this way, if we have enough information (3 of the 4 variables), we can solve for
the missing parameter by using one form of the equation.
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Example (Solving for Missing Parameters)
1. A student got a 580 on her verbal GRE, putting her at the lower 16th percentile. The report said that the standard deviation of verbal GRE scores
was 48. What was the mean of the scores? Use the Empirical Rule for this
problem.
2. In 2004, STAT 119 exam scores were normally distributed with a mean of 75
points. Unfortunately, the standard deviation and all scores were lost, but
one z-score of -2.15, which corresponded to a raw exam score of 60. What
was the standard deviation of these exam scores?
5.4
Inverse Normal Probability Calculations
In the previous sections, we were given a distribution and asked to find the
proportion of the distribution to the left of, right of, or between certain values
from the distribution. However, sometimes we will be given a region of the
normal curve, like “the lower 30%” and asked to find the value marking off that
region.
Given a distribution X ∼ N (µ, σ ), the general method for solving this type of
problem is as follows:
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
1. Draw a picture of the region.
2. On the z-table, look for the z-score corresponding to the percentile closest to
the one related to the area of interest. (e.g. For the lower 30%, look in the
middle of the table for the number closest to .3000 and choose its z-score).
3. Use the equation x = zσ + µ to unstandardize your z-score into a value
from X.
Example (Inverse Probability Calculations)
1. In the standard normal model, what value(s) of z cut(s) off the region described?
(a) The lowest 10%
(b) The highest 30%
(c) The middle 70%
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2. Polygraph machines rely in part on galvanic skin response, or the skin’s
natural ability to conduct electricity, to assess a person’s emotional state.
Suppose that skin response scores are approximately normal with a mean
of 49.40 and a standard deviation of 3.00 (units unknown).
Suspects with skin response scores in the upper 3% are thought to be lying.
Find the minimum score that indicates lying.
3. Suppose (old) SAT verbal scores are normal with mean 500 and standard
deviation 100. Find the two test scores representing the middle 60%.
4. In MathXL, there are questions telling you
µ = 20,
25% above 22,
σ=
Indicating that a distribution is centered at 22 with 25% of the distribution
lying above 22. Find the missing parameter σ.
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CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)
5. A machine is used to regulate the amount of dye dispensed for mixing
shades of paint. The amount of dye per can follows a normal distribution
with σ = 0.4. If more than 6 mL of dye is discharged while making paint,
the shade is unacceptable. Determine the mean amount of paint discharged
(µ) so that only 1% of the cans of paint will be unacceptable.
5.5
Preparation for the Quiz
The reading for this lecture is Chapter 6. Normal Probability Plots (pg. 141-142)
will not be covered in class, but may be required knowledge for the quiz and
exam.
Practice Problems
Chapter 6:
32, 33, 39–50, 52, 55, 60
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