the normal probability distribution points of inflection 2 3 2 3 2 3 2 3
TRANSCRIPT
The Normal Probability Distribution
Points of Inflection
2 3 2 3
2 3 2 3
Main characteristics of the Normal Distribution
• Bell Shaped, symmetric
• Points of inflection on the bell shaped curve are at – and + That is one standard deviation from the mean
• Area under the bell shaped curve between – and + is approximately 2/3.
• Area under the bell shaped curve between – 2 and + 2is approximately 95%.
• Close to 100% of the area under the bell shaped curve between – 3 and + 3
There are many Normal distributions
depending on by and
0
0.01
0.02
0.03
0 50 100 150 200
x
f(x)
0
0.01
0.02
0.03
0 50 100 150 200
x
f(x)
0
0.01
0.02
0.03
0 50 100 150 200
x
f(x)
Normal = 100, = 40 Normal = 140, =20
Normal = 100, =20
The Standard Normal Distribution = 0, = 1
0
0.1
0.2
0.3
0.4
-3 -2 -1 0 1 2 3
• There are infinitely many normal probability distributions (differing in and )
• Area under the Normal distribution with mean and standard deviation can be converted to area under the standard normal distribution
• If X has a Normal distribution with mean and standard deviation than has a standard normal distribution
has a standard normal distribution.
• z is called the standard score (z-score) of X.
X
z
Converting Area
under the Normal distribution with mean and standard deviation
to
Area under the standard normal distribution
Perform the z-transformation
then
Area under the Normal distribution with mean and standard deviation
X
z
P a X b
a X bP
a bP z
Area under the standard normal distribution
P a X b
Area under the Normal distribution with mean and standard deviation
a b
a bP z
Area under the standard normal distribution
a b
0
1
Using the tables for the Standard Normal distribution
Example
Find the area under the standard normal curve between z = - and z = 1.45
9265.0)45.1( zP
• A portion of Table 3:
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06
1.4 0.9265
...
...
0 145.
9265.0
z0 145.
9265.0
z
P z( 0. ) 98 .01635
Example
Find the area to the left of -0.98; P(z < -0.98)
0000.98
Area asked for
0.980.98
Area asked for
0735.09265.00000.1)45.1( zP
Example
Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)
9265.0
145.
Area asked for
0 z
9265.0
145.
Area asked for
145.
Area asked forArea asked for
0 z0 z
4265.05000.09265.0)45.1( zP
Example
Find the area to the between z = 0 and of z = 1.45; P(0 < z < 1.45)
• Area between two points = differences in two tabled areas
145.0 z145.145.0 z0 z
Notes
Use the fact that the area above zero and the area below zero is 0.5000
the area above zero is 0.5000
When finding normal distribution probabilities, a sketch is always helpful
3962.01038.05000.0)026.1( zP
Example:Find the area between the mean (z = 0) and z = -1.26
0 z 1 2 6.
A r e a a s k e d f o r
0 z0 z 1 2 6.
A r e a a s k e d f o r
1 2 6.
A r e a a s k e d f o rA r e a a s k e d f o r
Example: Find the area between z = -2.30 and z = 1.80
9534.00107.09641.0)80.126.1( zP
0 .. - 2 . 3 0
Area Required
1 . 8 00 .. 0 .. 0 .. - 2 . 3 0
Area Required
1 . 8 0- 2 . 3 0
Area Required Area Required
1 . 8 0
Example: Find the area between z = -1.40 and z = -0.50
2277.00808.03085.0)50.040.1( zP
Area asked for
-1.40 -0.500
Area asked for
-1.40 -0.50
Area asked for
-1.40 -0.5000
Computing Areas under the general Normal Distributions
(mean , standard deviation )
1. Convert the random variable, X, to its z-score.
Approach:
3. Convert area under the distribution of X to area under the standard normal distribution.
2. Convert the limits on random variable, X, to their z-scores.
X
z
b
za
PbXaP
Example
Example: A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally
distributed and a bottle is selected at random:
1) Find the probability the bottle contains between 32.00 oz and 32.025 oz
2) Find the probability the bottle contains more than 31.97 oz
When x z 32.0032.00 32.00 32.0
0.00;0.02
Solutions part 1)
When x z
32 025
32.025 32 025 32.0125. ;
.
0.02.
P X PX
P z
( . )0. 0.
.
0.
( . ) .
32.0 32 02532.0 32.0
02
32.0
02
32 025 32.0
02
0 125 0 3944
Graphical Illustration:
3 2 . 0 x0 1 2 5. z
3 2 0 2 5.
A r e a a s k e d f o r
3 2 . 0 x3 2 . 0 x0 1 2 5. z0 1 2 5. z
3 2 0 2 5.
A r e a a s k e d f o r
3 2 0 2 5.
A r e a a s k e d f o r
P x Px
P z( . ).
( .
. . .
319732.0
0.023197 32.0
0.02150)
1 0000 00668 0 9332
Example, Part 2)
32.03197. x
0150. z32.03197. x
0150.150. z
Combining Random Variables
Quite often we have two or more random variables
X, Y, Z etc
We combine these random variables using a mathematical expression.
Important question
What is the distribution of the new random variable?
An Example
Suppose that a student will take three tests in the next three days
1. Mathematics (X is the score he will receive on this test.)
2. English Literature (Y is the score he will receive on this test.)
3. Social Studies (Z is the score he will receive on this test.)
Assume that
1. X (Mathematics) has a Normal distribution with mean = 90 and standard deviation = 3.
2. Y (English Literature) has a Normal distribution with mean = 60 and standard deviation = 10.
3. Z (Social Studies) has a Normal distribution with mean = 70 and standard deviation = 7.
Graphs
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0 20 40 60 80 100
X (Mathematics) = 90, = 3.
Y (English Literature) = 60, = 10.
Z (Social Studies) = 70 , = 7.
Suppose that after the tests have been written an overall score, S, will be computed as follows:
S (Overall score) = 0.50 X (Mathematics) + 0.30 Y (English Literature) + 0.20 Z (Social Studies) + 10 (Bonus marks)
What is the distribution of the overall score, S?
Sums, Differences, Linear Combinations of R.V.’s
A linear combination of random variables, X, Y, . . . is a combination of the form:
L = aX + bY + …
where a, b, etc. are numbers – positive or negative.
Most common:Sum = X + Y Difference = X – Y
Others
Averages = 1/3 X + 1/3 Y + 1/3 Z
Weighted averages = 0.40 X + 0.25 Y + 0.35 Z
Means of Linear Combinations
The mean of L is:
Mean(L) = a Mean(X) + b Mean(Y) + …
L = a X + b Y + …
Most common:
Mean( X + Y) = Mean(X) + Mean(Y)
Mean(X – Y) = Mean(X) – Mean(Y)
If L = aX + bY + …
Variances of Linear Combinations
If X, Y, . . . are independent random variables and
L = aX + bY + … then
Variance(L) = a2 Variance(X) + b2 Variance(Y) + …
Most common:
Variance( X + Y) = Variance(X) + Variance(Y)
Variance(X – Y) = Variance(X) + Variance(Y)
2 2 2 2 2L X Ya b
If X, Y, . . . are independent normal random variables, then L = aX + bY + … is normally distributed.
In particular:
X + Y is normal with
X – Y is normal with
Combining Independent Normal Random Variables
22 deviation standard
mean
YX
YX
22 deviation standard
mean
YX
YX
Example: Suppose that one performs two independent tasks (A and B):
X = time to perform task A (normal with mean 25 minutes and standard deviation of 3 minutes.)
Y = time to perform task B (normal with mean 15 minutes and std dev 2 minutes.)
X and Y independent so T = X + Y = total time is normal with
6.323 deviation standard
401525 mean
22
0823.39.16.3
404545
ZPZPTP
What is the probability that the two tasks take more than 45 minutes to perform?
The distribution of averages (the mean)
• Let x1, x2, … , xn denote n independent random variables each coming from the same Normal distribution with mean and standard deviation .
• Let
11 2
1 1 1
n
ii
n
xx x x x
n n n n
What is the distribution of ?x
The distribution of averages (the mean)
Because the mean is a “linear combination”
1 2
1 1 1nx x x xn n n
and
1 1 1 1n
n n n n
1 2
2 2 22 2 2 21 1 1
nx x x xn n n
2 2 2 2 22 2 2
2
1 1 1n
n n n n n
Thus if x1, x2, … , xn denote n independent random variables each coming from the same Normal distribution with mean and standard deviation .
Then
11 2
1 1 1
n
ii
n
xx x x x
n n n n
has Normal distribution with
mean andx 2
2variance x n
standard deviation xn
Example
• Suppose we are measuring the cholesterol level of men age 60-65
• This measurement has a Normal distribution with mean = 220 and standard deviation = 17.
• A sample of n = 10 males age 60-65 are selected and the cholesterol level is measured for those 10 males.
• x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, are those 10 measurements
Find the probability distribution of
Compute the probability that is between 215 and 225
?xx
Example
• Suppose we are measuring the cholesterol level of men age 60-65
• This measurement has a Normal distribution with mean = 220 and standard deviation = 17.
• A sample of n = 10 males age 60-65 are selected and the cholesterol level is measured for those 10 males.
• x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, are those 10 measurements
Find the probability distribution of
Compute the probability that is between 215 and 225
?xx
Solution
Find the probability distribution of x
Normal with 220x 17
and 5.37610
xn
215 225P x
215 220 220 225 220
5.376 5.376 5.376
xP
0.930 0.930 0.648P z
Graphs
0
0.02
0.04
0.06
0.08
150 170 190 210 230 250 270 290 310
The probability distribution of individual observations
The probability distribution of the mean
Normal approximation to the Binomial distribution
Using the Normal distribution to calculate Binomial probabilities
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
-0.5
Binomial distribution
Approximating
Normal distribution
Binomial distribution n = 20, p = 0.70
049.2
14
npq
np
Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n and p
21
21 aYaPaXP
• Y has a Normal distribution
npq
np
correction continuity21
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
Binomial distribution
-
0.0500
0.1000
0.1500
0.2000
0.2500
a
-
-0.5
Approximating
Normal distribution
P[X = a]
21a 2
1a
-
0.0500
0.1000
0.1500
0.2000
0.2500
a-
-0.5
21
21 aYaP
-
0.0500
0.1000
0.1500
0.2000
0.2500
a
-
-0.5
P[X = a]
Example
• X has a Binomial distribution with parameters n = 20 and p = 0.70
13 want We XP
13 eexact valu The XP
1643.030.070.013
20 713
Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:
049.230.70.20
14)70.0(20
npq
np
21
21 131213 YPXP
Hence
5.135.12 YP
049.2
145.13
049.2
14
049.2
145.12 YP
= 0.4052 - 0.2327 = 0.1725
24.073.0 ZP
Compare with 0.1643
Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n and p
21
21 bYaP
• Y has a Normal distribution
npq
np
correction continuity21
)()1()( bpapapbXaP
-
0.0500
0.1000
0.1500
0.2000
0.2500
a b
-
-0.5
21a 2
1b
bXaP
-
0.0500
0.1000
0.1500
0.2000
0.2500
a b
-
-0.5
21a 2
1b
21
21 bYaP
Example
• X has a Binomial distribution with parameters n = 20 and p = 0.70
1411 want We XP 1411 eexact valu The XP
614911 30.070.014
2030.070.0
11
20
)14()13()12()11( pppp
5357.01916.01643.01144.00654.0
Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:
049.230.70.20
14)70.0(20
npq
np
21
21 14101411 YPXP
Hence
5.145.10 YP
049.2
145.14
049.2
14
049.2
145.10 YP
= 0.5948 - 0.0436 = 0.5512
24.071.1 ZP
Compare with 0.5357
Comment:
• The accuracy of the normal appoximation to the binomial increases with increasing values of n
Example• The success rate for an Eye operation is 85%• The operation is performed n = 2000 times
Find1. The number of successful operations is
between 1650 and 1750.2. The number of successful operations is at
most 1800.
Solution
• X has a Binomial distribution with parameters n = 2000 and p = 0.85
17201680 want We XP
5.17205.1679 YP
where Y has a Normal distribution with:
969.1515.85.200
1700)85.0(2000
npq
np
17201680 Hence XP
969.15
17005.1720
969.15
1700
969.15
17005.1679 YP
= 0.9004 - 0.0436 = 0.8008
28.128.1 ZP
5.17205.1679 YP
Solution – part 2.
1800 want We XP
5.1800 YP
969.15
17005.1800
969.15
1700YP
= 1.000
29.6 ZP