Download The Normal Curve

REVIEW PROBLEM
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
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.
2.
3.
4.
State the problem.
Use the table.
Draw & shade a picture.
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:
z
x x
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.
795 490
z
100
305
100
3.05
In doing so, we see that this student scored 3.05 standard
deviations above the mean of 490.
The Standard Normal Distribution
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.
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
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?
16.2 15.8
z
0.6
0.4
0.6
0.6667
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.
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
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.
90 72.3
z
8.9
82 72.3
z
8.9
17.7
1.9888
8.9
9.7
1.0899
8.9
We get 0.9767 and 0.8621 as the probability associated with these 2 z-scores.
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
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.
We get 0.9767 and 0.8621 as the probability associated with these 2 z-scores.
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.
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?
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?
75 50
z
10
25
10
2.5
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.
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.
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?
37 50
10
13
1.3
10
This yields a proportion of
0.0968 to the left.
z
Therefore, 9.68% of
all aliens should be
expected to serve
their duty on the front
lines of battle.
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.
3. What percent of all aliens are between 41 and 71 cm tall?
z
71 50
10
21
2.1
10
This yields proportions to the left of 0.9821 and 0.1841.
z
41 50
10
9
10
0.9
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.