chapter 5 the normal distribution (pt. 2) · pdf fileelem. business statistics chapter 5. the...

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 thebottom N% of the distribution. For review, remember that Q1 is the 25th per-centile, the median is the 50th percentile, and Q3 is the 75th percentile. A normalpercentile is simply a percentile of a normally-distributed variable.

In the last lecture, we learned about the empirical rule’s estimates for the normaldistribution’s central regions. Let us extend that rule to make some estimateson normal percentiles. Let X ∼ N(µ, σ). Then, by application of the empiricalrule:

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 gotthese numbers.

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Elem. Business Statistics CHAPTER 5. THE NORMAL DISTRIBUTION (PT. 2)

Note that if x marks the Nth percentile, then it follows straightforwardly that xalso marks off the top (100− N)% of the distribution.

The table of standard normal percentiles or, more colloquially, the z-table, per-centile information on the standard normal distribution. Thus, the percentiles inthis case are actually z-scores. The table ranges from −3.90 ≤ z ≤ 3.90 in incre-ments 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 corre-sponds by first looking down the left margin and locating 1.1. You then movealong 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 · · · 0.05 · · · 0.09

0.0...

......

......

......

......

......

1.1 · · · · · · 0.8749 · · · · · ·...

......

......

...

3.9...

......

......

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% ofthe standard normal curve lies to the left of z = 1.15, it follows that if we wereto 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). Alsonote that, for technical reasons, P(Z < z) = P(Z ≤ z). That is, including z doesnot make a difference.

Now, because any normal distribution (e.g. X ∼ N(µ, σ)) can be converted to thestandard normal distribution (e.g. Z ∼ N(0, 1)), we can take any value x from

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X, convert it to a z-score, and then use the z-table to find out what percentilex corresponds to in its original distribution. Therein lies the usefulness of thestandard normal distribution.

Example (Standard Normal Percentiles)

For the following, we use the model Z ∼ N(0, 1). For each problem, you shouldfirst 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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Example (Non-Standard Normal Percentiles)

For the following, we use the distribution X ∼ N(1152 pounds, 84 pounds) de-scribing the weights of young Angus steers in Problem 5.19. As before, draw apicture of the model and shade in the area of interest. Also, note that the z-tableonly gives us z-values up to the second decimal place, so round accordingly.


1. The percentage of Angus steers weighing less than 1250 pounds.

2. The probability of randomly choosing an Angus steer weighing less than1000 pounds.

3. If a z-score is a measure of unusualness, then which would be more un-usual: 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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5.2 Using Normal Percentiles to Find More GeneralAreas of the Normal Curve

The standard normal percentiles in the z-table give the proportion of the normalcurve less than, or to the left of, a given z-value. However, we might be interested inthe areas of more general areas of the standard normal curve, like the proportiongreater 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 per-centiles. We name them Rule L(ess Than), Rule G(reater Than), and RuleB(etween):

Rule L: Let z be a z-value. Then the proportion of the standard normal curveless than z is simply its entry in the z-table. That is:

P(Z < z) = Entry in the z-table

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

x

dnorm(x)

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Rule G: Let z be as before. The proportion of the standard normal curve greaterthan z is 1 minus z’s entry in the table. That is:

P(Z > z) = 1− Entry in the z-table = 1− P(Z < z)

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

x

dnorm(x)

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

x

dnorm(x)

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Rule B: Let z1 < z2 be two z-scores. The proportion of the standard normalcurve between z1 and z2 is the difference between the areas left of z2 and z1.That is:

P(z1 < Z < z2) = Entry for z2 − Entry for z1 = P(Z < z2)− P(Z < z1)

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

x

dnorm(x)

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

x

dnorm(x)

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

x

dnorm(x)

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Example (Areas of the Standard Normal Curve)

For the following, use the standard normal model (Z ∼ N(0, 1)). As before, drawa 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.


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 othervariables on the left hand side:

x = zσ + µ

σ =x− µ

zµ = x− zσ

In this way, if we have enough information (3 of the 4 variables), we can solve forthe 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 per-centile. The report said that the standard deviation of verbal GRE scoreswas 48. What was the mean of the scores? Use the Empirical Rule for thisproblem.

2. In 2004, STAT 119 exam scores were normally distributed with a mean of 75points. Unfortunately, the standard deviation and all scores were lost, butone z-score of -2.15, which corresponded to a raw exam score of 60. Whatwas 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 theproportion of the distribution to the left of, right of, or between certain valuesfrom the distribution. However, sometimes we will be given a region of thenormal curve, like “the lower 30%” and asked to find the value marking off thatregion.

Given a distribution X ∼ N(µ, σ), the general method for solving this type ofproblem is as follows:

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1. Draw a picture of the region.

2. On the z-table, look for the z-score corresponding to the percentile closest tothe one related to the area of interest. (e.g. For the lower 30%, look in themiddle 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 valuefrom X.

Example (Inverse Probability Calculations)

1. In the standard normal model, what value(s) of z cut(s) off the region de-scribed?

(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’snatural ability to conduct electricity, to assess a person’s emotional state.Suppose that skin response scores are approximately normal with a meanof 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 standarddeviation 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 distributionlying above 22. Find the missing parameter σ.

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5. A machine is used to regulate the amount of dye dispensed for mixingshades of paint. The amount of dye per can follows a normal distributionwith 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 andexam.

Practice Problems

Chapter 6:32, 33, 39–50, 52, 55, 60

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chapter 5 the normal distribution (pt. 2) · pdf fileelem. business statistics chapter 5. the...

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