Download 2.2 Normal Distributions

2.2 Normal Distributions
Normal curves which describe Normal Distributions
are anything but normal!
These curves are special!
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Some attributes of Normal Curves
1. All Normal curves have the same overall shape:
a) Symmetric, single­peaked, bell­shaped
2. Any specific Normal curve is completely described by giving its mean μ and its standard deviation σ.
μ
σ
μ σ
The mean is located at the center of the symmetric curve and is the same as the median.
The standard deviation controls the spread of a Normal curve.
We can eyeball the σ by looking for the inflection point of the curve.
Calculus is Alive!
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Some Basic Properties of the Normal Distribution
1. Centered about the Mean = μ (Mu)
2. The Standard Deviation = σ (Sigma) ­Describes how spread out the Normal Distribution.
3. Also referred to as a Density Curve
4. The Total area under the curve is 1 or another way of looking at is the Total Probability = 1 or 100%
5. The curve is symmetric about the mean.
6. The two tails of the curve extend Indefinitely!
­ b/c a Normal Random variable can take on any value ( ­ inf, inf )
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Normal Distributions
1. Different Means ­ Location changes but the spread is the same so change the μ then just change the location of your Normal Standard curve. (St. Dev. are the same)
μ 1
μ2
μ3
2. Different Standard Deviations but same Mean
a) Location is the same but the spread is different.
μ
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Definition:
Normal distribution and Normal curve
A Normal distribution is described by a Normal density curve. Any particular Normal distribution is competely specified by two numbers; its mean μ and standard deviation σ. The mean of a Normal distribution is at the center of the symmetric Normal curve. The standard deviation is the distance from the center to the change­of­
curvature points on either side.
We abbreviate the Normal distribution with mean μ and the standard deviation σ as N(μ,σ).
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THREE REASONS THE NORMAL CURVE IS IMPORTANT TO STATISTICS.
1. Normal distributions are good descriptions for some distributions of real data.
a. scores on tests taken by many people (SAT, IQ)
b. repeated careful measurements of the same quantity (like diameter of a tennis ball)
c. characteristics of biological populations 2. Normal distributions are good approximations to the results of many chance outcomes, like the number of heads in many tosses of a fair coin.
3. Many statistical inference procedures are based on Normal distributions. ( this is the most important)
Caution: Even though many data sets follow a Normal distribution, many do not!
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Normal Curve Applet activity p111
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The 68­95­99.7 Rule
Definition:
The 68­95­99.7 Rule
Better Know This
In the Normal distribution with mean μ and standard deviation σ
Approximately 68% of the observations fall within σ of the μ.
Approximately 95% of the observations fall within 2σ of the μ.
Approximately 99.7% of the observations fall within 3σ of the μ.
Example: P113
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Example: ITBS Vocabulary Scores
Using the 68­95­99.7 Rule
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Chebyshev's Inequality (Not on the AP Test)
The 68­95­99.7 Rule only applies to Normal distributions. Chebyshev's Inequality says that in any distribution the proportion of observations falling within k standard deviations of the mean is at least 1­1/k2
For example: If k = 2 then the inequality tells us that at least (1­1/22) = .75 of the observations in any distributions are within 2 standard deviations of the mean.
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THE STANDARD NORMAL DISTRIBUTION
Definition:
Standard Normal Distribution
The standard Normal distribution is the Normal distribution with mean 0 and standard deviation 1.
If a variable x has any Normal distribution N(μ,σ) with mean μ and standard deviation σ, then the standardized variable
z = (x ­ μ)/σ has the standard Normal distribution.
(z = (x ­xbar)/s for sample data)
Notes: 1. All Normal distributions are the same when we standardize.
2. 68­95­99.7 can only be used for the 68% which is z = 1 or ­1.
3. We use a table to find areas under the curve for the standard Normal distributions.
z ­ scores ­ How many standard deviations an observation from the standard Normal Distribution lies from the mean.
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Definition: The standard Normal table
Table A is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z.
See page 115 ­ 116
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Example page 116 and 117
Standard Normal Distribution p116
z
­1.8
­1.7
­1.6
.07
.0307
.0384
.0475
.08
.0301
.0375
.0465
.09
.0294
.0367
.0455
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P117 Catching Some Z's
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Technology corner p118
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Ex. Finding the standardized score for the following:
Find the z ­ score for 33
Data: 24 28
Mean: 39
33
33
37
39
47
51
59
Standard Deviation: 11.34681
z = (33 ­ 39)/11.34681 = ­.53
33 lies .53 standard deviations below the mean (below b/c it was negative)
Recall: How did I calculate the standard deviation?
=sqrt((225+121+36+36+4+0+64+144+400)/(9 ­ 1))
sqrt(1030/8)
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Example: Data: 3
5
6
Mean: 8.916667
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7 9 9 10
11
12
14
15
St. Dev.: 3.679386
Find the Z­Score of 12
z = (12 ­ 8.916667)/3.679286 = .84
12 lies .84 Standard deviations above the Mean. 17
Example: The following are high temperatures fro 8 days in Jan. for Chicago in degrees F.
1
6
16
18
24
25
25
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a) Find the Quartiles
b) Find the 30th %tile
c) Find the z ­ score for 25
solution:
a) (8 observations)
Q2 = (18+24)/2 = 21
Q1 = (6+16)/2 = 11
Q3 = (25+25)/2 = 25
b) k = 30
n = 8
30(8)/100 = 2.4 P30 ≈16
c) Mean = 18.25
3rd observation (round up)
St. Dev. = 10.27827
z = (25­18.25)/10.27827 = .66
25 lies .66 St. Dev. above the mean.
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Normal Distribution Calculations
How to Solve Problems Involving Normal Distributions
1. STATE: Express the problem in terms of the observed variable x
2. PLAN: Draw a picture of the distribution and shade the area of interest under the curve.
3. DO: Perform calculations.
a) Standardize x to restate the problem in terms of a standard Normal variable z.
b) Use Table A and the fact that the total area under the curve is 1 to find the required area under the standard Normal curve.
4. CONCLUDE: Write your conclusion in the context of the problem.
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Do all examples from pages 120 ­ 124
Tiger on the Range
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Tiger on the Range cont. P121
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Cholesterol in Young Boys
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Technology Corner: From z­scores to areas, and vice versa
p123
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Ex: Find the area under the Standard Normal Curve to the left of z = 1.95
1. draw pic
2. The chart in book give you the values to the left = .9744
Which is 97.44% of the values lie to the left of 1.95
Ex2: Find the area under the standard Normal Curve from z = ­2.17 to z = 0
1. This is the area between the 2 points. Find the diff. between the Right bound and Left Bound. (z­score for (0 ­ (­2.17))
The table gives the values as:
.5 ­ (.0150) = .4850 so about 48.50% of the values lie in between those points.
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Find the Area under the Standard Normal Curve to the right of z= 2.32
Since the table gives values to the right we need to take 1 ­ z
1 since total is 1.
1 ­ (.9898) = .0102
So 1.02% of values lie above 2.32
Find the Area under the Standard Normal Curve to the right of ­1.54
So take total = 1 and subtract z of ­1.54
1 ­ .0618 = .9382
So 93.82% lie to the right of ­1.54
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These problems symbolically might look like:
P(1.19< z <2.12)
P(­1.56< z < 2.31)
P(z > ­.75)
See page 104 black book to see answers
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Special Case:
P (0 < z < 5.67)
5.67 is off the chart so area to the left of 5.67 is approximately 1
therefore 1 ­ .5 = .5
Remember zero is the middle so 50%
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Try these:
P(­2.07 < z < ­.93)
P (z > 1.48 )
See page 105 black book for solutions
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Standardizing a Normal Distribution
For a Normal random variable X, a particular value of x can be converted to its corresponding z­value by using the formula:
z = (x ­ μ)/σ
Necessary b/c chart is only in z­scores
Ex: Let x be a continuous random variable that has a Normal distribution with a mean of 50 and st. dev. of 10. Convert the following x values to z values and find the area to the left of these points.
1. x = 55
z = (55­50)/10 = .5
chart of .5 = .6915
2. x = 35
z = (35­50)/10 = ­1.5
chart of ­1.5 = .0668
More examples on p107­ 108 of black book
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Assessing Normality ­ p125
See Example p125 Unemployment in the States and Ex p126 Making a Normal probability plot
Unemployment in the States:
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Interpreting Normal Probability Plots
If the points on a Normal probability plot lie close to a straight line, the plot indicates that the data are Normal. Systematic deviations from a straight line indicate a non­
Normal distribution. Outliers appear as points that are far away from the overall pattern of the plot.
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Guinea Pig Survival: p127
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Technology Corner: p129
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Application of Normal Distribution
U.S. Debt
The average consumer debt of U.S. households was $17,989 in 2004. Suppose these debts have a Normal distribution with a μ = $17,989 and a σ = $3750. Find the probability that such debt for a randomly selected U.S. household is between $13000 and $20000.
μ = $17989 σ = $3750
P( 13000 < x < 20000)
z = (13000 ­ 17989)/3750 = ­1.33
z = (20000 ­ 17989)/3750 = .54
chart .54 ­ chart ­1.33
.7054 ­ .0918 = .6136
So about 61% of all U.S. households have debt in that range.
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Ex: 2
The assembly time of toy cars follows a Normal distribution with μ = 55 min. and σ = 4 min. The company closes at 5 p.m. every day. If one worker starts to assemble a car at 4 p.m., what is the probability that she will finish the job before the company closes for the day?
μ = 55 min
σ = 4
z = (60­55)/4 = 1.25
P(x < 60 min)
chart 1.25 = .8944
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Ex3: The filling machine of a soda dispenser pours liquid into a can. The amount follows a Normal distribution with μ = 12 ounces and σ = .015 ounces. What is the probability that a randomly selected can of soda contains 11.97 ounces to 11.99 ounces?
μ = 12 σ = .015
P(11.97 < x < 11.99)
z = (11.97 ­ 12)/.015 = ­2
z = (11.99 ­ 12)/.015 = ­.67
chart ­.67 ­ chart ­2
.2514 ­ .0228 = .2286
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Ex4: Life span of a calculator has a Normal distribution with an average of 54 months and a σ = 8 months. The company guarantees that any calculator that starts to malfunction within 36 months will be replaced. About what % of calculators are expected to be replaced?
μ = 54 months σ = 8 months
z = (36­54)/8 = ­2.25
P( x < 36)
chart ­2.25 = .0122
about1.22%
More Examples in Black book p111 ­ 116
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Homework:
p131 41­51odd
p132
53­59odd
p133
63, 65, 66, 68­74all
Unit 2 Review, Practice Test, Frappy and 10 MC
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