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Welcome to MM207 - Statistics!
Unit 5 Seminar:
Good Evening Everyone!
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Normal Probability Distribution
It is a continuous probability
distribution
Two values determine its shape
μ = mu
mean of distribution
σ = sigma
standard deviation
of the distribution
μ - 3σ
μ - 2σ
μ-σ
μ
μ+σ
μ + 2σ
μ + 3σ
Normal Probability Distribution
Several important properties
(page 240 text)
The curve is bell-shaped with the
highest point centered on µ
It is symmetrical about a vertical
line through µ
The curve approaches the
horizontal axis but never
touches it
The total area= 1.
μ - 3σ
μ - 2σ
μ-σ
μ
μ+σ
μ + 2σ
μ + 3σ
Standard Normal Curve
µ=0
σ =1
z score – tells us how standard
deviations away from the
mean a value is:
z = (x - µ)/ σ
We will convert x values
to z scores using
the above formula
[page 107 or 243] or Excel!
Example 1: Using the Table
P(z ≤ 1.28) = ?
Using the Standard Normal Table
look up the area associated with z = 1.28
Note: The table only gives areas to the
left of the given z score.
Find the row for z = 1.2
Find the column for 0.08
The area is 0.8997
P(z ≤ 1.28) = 0.8997
-3
-2
-1
0
1 1.28
This area makes sense because the area needed
is more than half of the area under the curve as shown in the sketch.
2
3
Example 1 cont’
P(z ≤ 1.28) = 0.8997
Excel can also be used to find the probability, the function is NORMSDIST.
The instructions are in Doc Sharing.
Example 2: Using the Table
P(z ≥ -0.56) = ?
Look up the area associated with -0.56 in
the table.
The table only gives the area to the left.
The area is 0.2877.
This is the area to the left of z = -0.56.
We want the area to the right.
P(z ≥ -0.56) =1 - 0.2877
= 0.7123
0.2877
-3
-2
Note: P(z ≥ - 0.56) is the same as P(z ≤ 0.56)
So you can also look up P(z ≤ 0.56) . Try it!
This is because the curve is symmetrical.
-1
0.7123
-0.56 0
1
2
3
Example 3: Using the Table
P(0 ≤ z ≤ 1) = ?
This value is not given in the table. We need to get creative.
P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0)
P(z ≤ 0) = 0.500
P(z ≤ 1) = 0.8413
P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0) = 0.8413 – 0.5000 = 0.3413
Always subtract the smaller area from the larger area. The answer
cannot be negative.
=
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
Example 3 cont’
P(z ≤ 0) = 0.500
P(z ≤ 1) = 0.8413
P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0) = 0.8413 – 0.5000 = 0.3413
Example 4: Finding a Probability for a given x
Given: µ = 600, σ =100
P(x ≥ 750) = ?
X distribution:
µ = 600
σ =100
300
400
500
CONVERT
600
700
800
750
900
Z distribution
µ=0
σ =1
-3
-2
-1
0
1
2
?
3
Example 4: Finding a Probability for a given x
Given: µ = 600, σ =100
P(x ≥ 750) = ?
Convert x to a z-score
z = (x - µ)/ σ
P(x ≥ 750) =
P(z ≥ (750 - 600)/ 100) = P(z ≥150/100)
P(z ≥ 1.5)
Use the table.
P(z ≥ 1.5) = 1 – P(z ≤ 1.5)
= 1 – 0.9332 = 0.0668
P(x ≥ 750) = 0.0668
You can also look up the opposite,
P(z ≤ - 1.5) = .0668
-3
-2
-1
0
1
Z = 1.5
2
3
Example 5: Working backwards
Find Z0 in P(z ≤ z0 ) = 0.99
Find the areas closest
to 0.9900
P(z ≤ 2.32 ) = 0.9898
P(z ≤ 2.33 ) = 0.9901
0.9901 is closer to 0.99
P(z ≤ 2.33 ) ≈ 0.99
z0 = 2.33
Example 6: Working Backwards
An IQ test has μ = 100 and σ = 15. To get into a special program, a
Student must have an IQ score in the top 10%. What is the lowest IQ
score needed?
Find the z score associated with 10%
Since we are looking for the top 10%, we use 90%
to find the z-score. 1 – 0.1000 = 0.9000
Find Z0 in P(z ≥ Z0 ) = 0.9000
Z0 = 1.28 (since 0.8997 is the area closest to 0.9000 in the table)
Convert to x
x = μ + zσ
x = 100 + 1.28 * 15 = 118.2
Round down to 118
The lowest IQ score needed is 118.
Sampling Distributions
and The Central Limit Theorem
Since we are often unable to measure the population mean and standard
deviation (μ and σ), we want to use the sample mean and standard deviation
(xbar and s) to estimate the parameters for the population.
One way we do this is by creating a sampling distribution from the population.
If we take all possible samples of a certain size from a population, and compute
the mean for each sample, we can create a sampling distribution of the means
The mean of the sampling distribution is the “mean of the means” which is μ.
μ xbar = μ
The standard deviation is the population standard deviation divided by the
square root of n.
σ xbar = σ/√n
Sampling Distributions
and The Central Limit Theorem (con’t)
The Central Limit Theorem (page 272) says that if we have samples
with size n ≥ 30 from the population, then the sampling distribution is
approximately normal even if the population distribution is not
normal.
If the population distribution is normal, then any sample size will be ok
and the sampling distribution will be normal.
This gives us the ability to find Confidence Intervals (Chapter 6) to
estimate the population mean and to find probabilities for hypothesis
testing (Chapter 7).
Example 7: Using Sampling Distributions
Scientists determined that x (the length of a single trout) is
normally distributed with µ = 10.2 and σ = 1.4
Question A:
What is the probability a single trout is between 8 and 12
inches? P(8 < x < 12) = ?
Question B:
What is the probability that the mean length of 5 trout
(xbar) will be between 8 and 12 inches?
P(8 < xbar < 12) = ?
Example 7 Solution
Given: µ = 10.2 and σ = 1.4 (normally distributed)
Question A: P(8 < x < 12) = ?
z = (x - µ)/σ
P(8 < x < 12)
= P((8 – 10.2)/1.4 < z < (12 – 10.2)/1.4)
= P(-1.57 < z < 1.29)
= P(z < 1.29) - P(z < -1.57) = 0.9015 - 0.0582 = 0.8433
Question B: P(8 < xbar < 12) = ?
z = (xbar - µxbar)/ σxbar
µxbar = µ = 10.2 and σxbar = σ/√n = 1.4/√5 ≈ 0.63
P(8 < xbar < 12)
= P((8 – 10.2)/0.63 < z < (12 – 10.2)/0.63)
= P(-3.49 < z < 2.86)
= P(z < 2.86) - P(z < -3.49) = 0.9979 - 0.0002 = 0.9977
Normal Approximations to Binomial Distributions
If np and nq are greater than or equal to 5, then we can use
the normal distribution to approximate a binomial
distribution. Page 285
A correction must be used, called the continuity correction.
Page 287
Move 0.5 units to the left and right of the midpoint to
include all possible x-values in the interval.
Then we can approximate a binomial probability
Example 8: Approximating Binomial Probabilities
Try it Yourself, page 280
What is the probability that at most (≤) 85 people will say yes?
n = 200
p = 0.38
np = 76 and nq = 124, so we can use the normal approximation
μ = 76; σ ≈ 6.86
Add 0.5 to 85 to get 85.5 for the continuity correction since this is x ≤ 85.
P( x ≤ 85.5) = P(z ≤ (85.5 – 76)/ 6.86) ) = P(z ≤ 1.38) = 0.9162