Residents of upstate New York are

accustomed to large amounts of snow with

snowfalls often exceeding 6 inches in one

day. In one city, such snowfalls were

recorded for two seasons and are as follows

(in inches):

8.6, 9.5, 14.1, 11.5, 7.0, 8.4, 9.0, 6.7, 21.5,

7.7, 6.8, 6.1, 8.5, 14.4, 6.1, 8.0, 9.2, 7.1

What are the mean and the population

standard deviation for this data, to the

nearest hundredth?

The mean = 9.46 inches.

Standard deviation = 3.74

REVIEW PROBLEM

Density Curves &

Normal Distribution

Some Call it a “Bell Curve”

DENSITY CURVES

Density curves give us a mathematical

model for distributions.

Density curves:

are always on or above horizontal axis

have an area of exactly 1 underneath it

Centers of Density Curves

MEDIAN—equal-areas point

MEAN—balance point

The two are equal for a

symmetric density curve.

The mean is pulled towards

the tail in a skewed curve.

In a skewed curve

Median = Equal Areas

Mean = Balance Point

NORMAL DISTRIBUTIONS

We use (mu) to represent the mean

and (sigma) to represent the standard

deviation of a normal distribution.

“Empirical Rule” 68-95-99.7 Rule

Normal Distribution

A normal distribution is a continuous, symmetrical,

bell-shaped distribution of a variable.

Characteristics of Normal Distribution

1. A normal distribution curve is bell-shaped.

2. The mean, median, and mode are equal and are located at the center of the distribution.

3. The curve is unimodal (i.e., it has only one mode)

4. The curve is symmetric about the mean (its shape is the same on both sides of a vertical line passing through the center)

5. The curve is continuous. There are no gaps or holes. For each value of X, there is a corresponding value of Y.

Characteristics of Normal Distribution

6. The curve never touches the x axis. No

matter how far in either direction the curve

extends, it never meets the x axis – but it

gets increasingly closer.

7. The total area under a normal distribution

curve is equal to 1.00, or 100%.

The area under the part of a normal curve that lies

within 1 standard deviation of the mean is

approximately 0.68, or 68%; within 2 standard

deviations, about 0.95, or 95%; and within 3

standard deviations, about 0.997, or 99.7%.

The Empirical Rule (a.k.a. the “68-95-99.7 Rule”)

In a normal distribution, almost all data will

lie within 3 standard deviations of the

mean.

o About 68% of all data lies within 1 standard

deviation of the mean.

o About 95% of all data lies within 2 standard

deviations of the mean.

o About 99.7% of all data lies within 3 standard

deviations of the mean.

Why Do We Need It?

The Empirical Rule is most often used in

statistics for forecasting or predicting final

outcomes.

After a standard deviation is calculated, and

before exact data can be collected, the Empirical

Rule can be used to estimate impending data.

The probability based on the Empirical Rule can

be used if gathering appropriate data may be

time consuming, or even impossible to obtain.

Tips for Using the Empirical Rule

Before applying the Empirical Rule it is a good idea to identify the data being described and the value of the mean and standard deviation.

Sketch a graph summarizing the information provided by the empirical rule and identify the percentages for each region of the graph (+/- 1, +/- 2, +/- 3)

Remember that data must be normally distributed for the Empirical Rule to apply.

The Empirical Rule

for a Normal Distribution

STEPS FOR FINDING

NORMAL PROPORTION

1. State the problem in terms of

x, your variable of concern.

2. Standardize the variable.

3. Draw a picture & shade area

of concern.

4. Use the table to determine

proportion.

From Proportion to Real

Value

GOING BACKWARDS!

1. State the problem.

2. Use the table.

3. Draw & shade a picture.

4. Unstandardize to obtain your x.

The amount of mustard dispensed from

a machine at The Hotdog Emporium

is normally distributed with a mean

of 0.9 ounce and a standard

deviation of 0.1 ounce. If the

machine is used 500 times,

approximately how many times will it

be expected to dispense 1 or more

ounces of mustard.

Choose:

5 16 80 100

The mean is 0.9 and the standard deviation is 0.1. If one standard deviation is

added to the mean, the result is 1.0 ounce. Therefore, dispensing 1 or more

ounces falls into the category above one standard deviation to the right of the

mean. Using the Empirical Rule, 16% of data falls at or above 1 standard

deviation.

16% x 500 = 80 times to dispense one or more ounces of mustard.

A machine is used to fill soda bottles. The amount of soda dispensed

into each bottle varies slightly. Suppose the amount of soda dispensed

into the bottles is normally distributed. If at least 99% of the bottles

must have between 585 and 595 milliliters of soda, find the greatest

standard deviation, to the nearest hundredth, that can be allowed.

The 99% implies a distribution within 3 standard deviations of the mean. The

difference from 585 milliliters to 595 milliliters is 10 milliliters. Symmetrically

divided, there are 5 milliliters used to create 3 standard deviations on one

side of the mean. Dividing 5 by 3, we get the standard deviation to be 1.67

milliliters, to the nearest hundredth.

Battery lifetime is normally distributed for large

samples. The mean lifetime is 500 days and

the standard deviation is 61 days. To the

nearest percent, what percent of batteries have

lifetimes less than 439 days?

Subtracting, we see that 1 s.d. below is 439 days, an exact match to

our question and an indication that the Empirical Rule can be used to

find the answer. The question is asking what percent of a distribution is

beyond one standard deviation to the left of the mean.

Answer: 16%

A shoe manufacturer collected data

regarding men's shoe sizes and found

that the distribution of sizes exactly fits

the normal curve. If the mean shoe size

is 11 and the standard deviation is 1.5,

find:

a. the probability that a man's shoe size

is greater than or equal to 11.

b. the probability that a man's shoe size

is greater than or equal to 14.

a. 50% In a normal distribution, the mean divides the data into two equal

areas. Since 11 is the mean, 50% of the data is above 11 and 50% is below 11.

b. 14 is exactly two standard deviations above the mean. Using the Empirical Rule

we see that 2.5% will fall above two standard deviations. Probability is 0.025.

Five hundred values are normally distributed with a mean of 125 and a standard

deviation of 10.

a. What percent of the values lies in the interval 115 - 135,

to the nearest percent?

b. What interval about the mean includes 95% of the data?

a. What percent of the values is in the interval 115 - 135?

mean + one standard deviation = 135

mean - one standard deviation = 115

Percent within one standard deviation of the mean = 68%

b. 2 standard deviations about the mean for a total interval size

of 40, with the mean in the center.

mean + 2 standard deviations = 145

mean - 2 standard deviations = 105

Interval: [105,145]

Example:

Estimating with the Empirical Rule

You have purchased fluorescent light

bulbs for your home. The average bulb life

is 500 hours with a standard deviation of

24. The data is normally distributed.

One of your bulbs burns out at 450 hours.

Would you send the bulb back for a

refund?

Problem: You have purchased fluorescent light bulbs for your home. The

average bulb life is 500 hours with a standard deviation of 24. The data is

normally distributed. One of your bulbs burns out at 450 hours. Would you send

the bulb back for a refund?

Solution: According to the Empirical Rule: o 68% of the light bulbs should last between

500 ± 24 or between 476 to 524 hours.

o 95% of the light bulbs should last between 500 ± 2(24) or between 452 to 548 hours.

o 99.7% of the light bulbs should last between 500 ± 3(24) or between 428 to 572 hours.

If the light bulb only lasted 450 hours, I would consider it a defective bulb. Less than 2.5% should last less than 452 hours.

Your Turn The scores for all high school seniors taking the verbal section of the Scholastic Aptitude Test (SAT) in a particular year had a mean of 490 and a standard deviation of 100. The distribution of SAT scores is normal.

1. What percentage of seniors scored between 390 and 590 on this SAT test?

2. One student scored 795 on this test. How did this student do compared to the rest of the scores?

3. A rather exclusive university only admits students who were among the highest 16% of the scores on this test. What score would a student need on this test to be qualified for admittance to this university?

1. What percentage of seniors scored

between 390 and 590 on this SAT test?

The mean = 490 and the standard deviation = 100. So, we can draw the Normal Curve to depict the data.

Calculate the difference

between the mean and

the test scores:

|390-490| = 100 1

s.d.

|590-490| = 100 1

s.d. So, about 68% of

seniors would score

between 390 and 590

on the SAT test.

2. One student scored 795 on this test. How did

this student do compared to the other scores?

Calculate the difference between the mean and the student’s test score:795-390 = 405

This student’s score is more than 3 std. deviations from the mean (3 standard deviations = 3 * 100)

Since about 99.7% of seniors

would score between 190 and

790 on the SAT test, this is an

exceptionally high score.

Only 0.15% of students would

have scores above 790. This

is an example of using the

Empirical Rule to estimate an

answer.

3 . A rather exclusive university only admits students

who were among the highest 16% of the scores on this

test. What score would a student need on this test to be

qualified for admittance to this university?

About 68% of the scores are between 390 and 590, so this leaves 32% of the scores outside this interval.

Since a bell-shaped curve is symmetrical, one-half of the scores (16%) are on each end of the distribution.

• 16% of the students scored above 590 on this SAT test.

So, to qualify for admittance to this university, a student would need to score 590 or above on the SAT test.

What if you want to be more precise?

The Empirical Rule works well if we are dealing

with values that are exactly 1, 2, or 3 standard

deviations away from the mean.

It also works well if we simply want to estimate

the probability or percentage for a particular

value.

However, we can be more precise using

NORMAL CALCULATIONS which involves

standardizing values to obtain z-scores.

STANDARD NORMAL

DISTRIBUTION

Standardizing yields z-scores.

Use this equation to find the

standardized value of x:

x xz

Standard Normal Curve

x has normal distribution of N( , )

z has a standard normal distribution N(0, 1)

The scores for all high school seniors taking the verbal section of the Scholastic Aptitude Test (SAT) in a particular year had a mean of 490 and a standard deviation of 100. The distribution of SAT scores is normal.

One student scored 795 on this test. How did this student do compared to the rest of the scores?

We estimated before that only 0.15% of students would score above 790, but what percent would score above 795?

First, we need to standardize the score of 795.

In doing so, we see that this student scored 3.05 standard deviations above the mean of 490.

3053.05

100

795 490100

z

The Standard Normal Distribution

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817

2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857

2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890

2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916

2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936

2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952

2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964

2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974

2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981

2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

This is part of a standard Normal table that you should be able to read. Since our

particular problem deals with a z-score of 3.05, we should go to the last row (3.0) and

over to the appropriate column (0.05). We should see that they meet at 0.9989. This is

the proportion TO THE LEFT of our score (of 795). Our question really wants us to

consider the proportion to the right. Since the total area under the curve is 1, we can

subtract 0.9989 from 1 to yield 0.0011. That means that only 0.11% (compared to

estimate of 0.15% from Empirical Rule) of all students scored higher than a 795.

A group of 625 students has a mean age

of 15.8 years with a standard deviation of

0.6 years. The ages are normally

distributed. How many students are

younger than 16.2 years? Express

answer to the nearest student?

0.40.6667

0.6

16.2 15.80.6

z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359

0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224

0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549

0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852

This z-score yields a percentage less than 16.2 as

74.86%. When we convert that to a number of students

we get 466.75 or approximately 467.

Professor Halen has 184 students in his

college mathematics lecture class. The scores

on the midterm exam are normally distributed

with a mean of 72.3 and a standard deviation

of 8.9. How many students in the class can be

expected to receive a score between 82 and

90? Express answer to the nearest student.

17.71.9888

8.9

90 72.38.9

z9.7

1.08998.9

82 72.38.9

z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441

1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545

1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633

1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

We get 0.9767 and 0.8621 as the probability associated with these 2 z-scores.

Professor Halen has 184 students in his college mathematics lecture

class. The scores on the midterm exam are normally distributed with a

mean of 72.3 and a standard deviation of 8.9. How many students in

the class can be expected to receive a score between 82 and

90? Express answer to the nearest student.

So, we can subtract 0.9767-0.8621 to yield a

proportion of 0.1146.

This means that 11.46% of all students scored

between an 82 and 90.

Therefore, roughly 21 (21.0864) of the 184

students scored between an 82 and 90.

We get 0.9767 and 0.8621 as the probability associated with these 2 z-scores.

1. Aliens place emphasis on height. For an alien to be a “leader” it would need to be at least 75 cm tall. What proportion of aliens would qualify?

2. Short aliens are often sent to the front lines in battle because they are smaller targets. All aliens that are under 37 cm tall can expect to be called into duty. What percent of aliens fall into this category?

3. What percent of all aliens are between 41 and 71 cm tall?

Aliens have come to visit Earth. We quickly establish that they are shorter than humans. In fact, their heights vary normally with a mean of 50 cm and standard deviation of 10 cm.

1. Aliens place emphasis on height. For an alien to be a “leader” it would need to be at least 75 cm tall. What proportion of aliens would qualify?

Aliens have come to visit Earth. We quickly establish that they are shorter than humans. In fact, their heights vary normally with a mean of 50 cm and standard deviation of 10 cm.

252.5

10

75 5010

z

This yields a proportion of 0.9938

to the left (or 0.0062 to the right).

So, only a proportion

of 0.0062 of all aliens

would qualify for a

leadership positions.

2. Short aliens are often sent to the front lines in battle because they are smaller targets. All aliens that are under 37 cm tall can expect to be called into duty. What percent of aliens fall into this category?

Aliens have come to visit Earth. We quickly establish that they are shorter than humans. In fact, their heights vary normally with a mean of 50 cm and standard deviation of 10 cm.

131.3

10

37 5010

z

This yields a proportion of

0.0968 to the left. Therefore, 9.68% of

all aliens should be

expected to serve

their duty on the front

lines of battle.

3. What percent of all aliens are between 41 and 71 cm tall?

212.1

10

71 5010

z

Aliens have come to visit Earth. We quickly establish that they are shorter than humans. In fact, their heights vary normally with a mean of 50 cm and standard deviation of 10 cm.

90.9

10

41 5010

z

This yields proportions to the left of 0.9821 and 0.1841.

So, when we subtract these

proportions, we find that the

proportion of aliens between

41 and 71 cm tall is 0.7980.

In other words, approximately

79.80% of all aliens are

between 41 and 71 cm tall.