Download The NORMAL DISTRIBUTION - Brocklehurst-13SAM

Statistics and Modelling
Course
2011
Topic 5: Probability Distributions
Achievement Standard 90646
Solve Probability Distribution
Models to solve straightforward
problems
4 Credits
Externally Assessed
NuLake Pages 278  322
PART 1
Lesson 1
• Overview of probability distributions (discrete
and continuous).
• The Standard Normal Distribution.
• HANDOUT: Just the first page of the Normal Distribution
handouts – on the std. normal distribution.
• HW:
– 1. In NuLake: Do pages 296 & 297 (intro to Normal Distn).
– 2. In Sigma – match-up task Q57: In OLD edition this is on
p98 (Ex. 7.1). In NEW edition it’s on p440 (exercise A.01).
Discrete Probability Distributions:
Discrete Probability Distributions:
Discrete Data is usually modelled using a Bar Graph with
Frequency on the vertical axis.
Discrete Probability Distributions:
Discrete Data is usually modelled using a Bar Graph with
Frequency on the vertical axis.
Probability = Frequency
Total
Continuous Probability Distributions:
Discrete Probability Distributions:
Discrete Data is usually modelled using a Bar Graph with
Frequency on the vertical axis.
Probability = Frequency
Total
Continuous Probability Distributions:
Continuous Data is modelled by a Histogram (using grouped
values) or a Probability Curve.
Discrete Probability Distributions:
Discrete Data is usually modelled using a Bar Graph with
Frequency on the vertical axis.
Probability = Frequency
Total
Continuous Probability Distributions:
Continuous Data is modelled by a Histogram (using grouped
values) or a Probability Curve. A bar graph is not appropriate
because continuous data contains no gaps!
Discrete Probability Distributions:
Discrete Data is usually modelled using a Bar Graph with
Frequency on the vertical axis.
Probability = Frequency
Total
Continuous Probability Distributions:
Continuous Data is modelled by a Histogram (using grouped
values) or a Probability Curve. A bar graph is not appropriate
because continuous data contains no gaps!
Probabilities are for a domain of values. A person is very
unlikely to be exactly 170.0…cm tall, or exactly 17 years old.
No limit to precision.
Discrete Probability Distributions:
Discrete Data is usually modelled using a Bar Graph with
Frequency on the vertical axis.
Probability = Frequency
Total
Continuous Probability Distributions:
Continuous Data is modelled by a Histogram (using grouped
values) or a Probability Curve. A bar graph is not appropriate
because continuous data contains no gaps!
Probabilities are for a domain of values. A person is very
unlikely to be exactly 170.0…cm tall, or exactly 17 years old.
No limit to precision.
Instead we’d ask for, say: P(169.5cm<X<170.5cm)
Continuous Probability Distributions:
Continuous Data is modelled by a Histogram (using grouped
values) or a Probability Curve. A bar graph is not appropriate
because continuous data contains no gaps!
Probabilities are for a domain of values. A person is very
unlikely to be exactly 170.0…cm tall, or exactly 17 years old.
No limit to precision.
Instead we’d ask for, say: P(169.5cm<X<170.5cm)
In this topic, we will use 3 types of probability distribution.
1.The Normal Distribution
2. The Binomial Distribution
3. The Poisson Distribution
Continuous Probability Distributions:
Continuous Data is modelled by a Histogram (using grouped
values) or a Probability Curve. A bar graph is not appropriate
because continuous data contains no gaps!
Probabilities are for a domain of values. A person is very
unlikely to be exactly 170.0…cm tall, or exactly 17 years old.
No limit to precision.
Instead we’d ask for, say: P(169.5cm<X<170.5cm)
In this topic, we will use 3 types of probability distribution.
FOR CONTINUOUS DATA:
1.The Normal Distribution
FOR DISCRETE DATA:
2. The Binomial Distribution
3. The Poisson Distribution
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for ____________ data (hence it is a _________curve).
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a ___________curve).
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures _____________,
NOT probability.
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution.
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution.
E.g. ____________________________________________
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution.
E.g. height, weight, dimensions of the human body.
2 Parameters:
__: _______
__: _______ ________
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: _______
__: _______ ________
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
__: _______ ________
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: _______ ________
The NORMAL DISTRIBUTION
 “Bell-shaped Curve”
 Used for continuous data (hence it is a continuous curve).
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution – symmetrical, greatest frequency is in the middle.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Cannot calculate probabilities for specific values because ____
_______________________________________________.
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution – symmetrical, greatest frequency is in the middle.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Cannot calculate probabilities for specific values because the
distribution is continuous – infinite degree of precision.
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution – symmetrical, greatest frequency is in the middle.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Cannot calculate probabilities for specific values because the
distribution is continuous – infinite degree of precision.
 We calculate probabilities for a _________ ___ _________.
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution – symmetrical, greatest frequency is in the middle.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Cannot calculate probabilities for specific values because the
distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. ____________________________________________.
 Height of the curve (vertical axis) measures FREQUENCY,
NOT probability.
 Measurements which occur in nature often fit a Normal
Distribution – symmetrical, greatest frequency is in the middle.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Cannot calculate probabilities for specific values because the
distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. Year 13s in NZ whose height is between 170 and 180cm.
 Measurements which occur in nature often fit a Normal
Distribution – symmetrical, greatest frequency is in the middle.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Cannot calculate probabilities for specific values because the
distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. Year 13s in NZ whose height is between 170 and 180cm.
PROBABILITY
=
______ ______ ___ _____
 Measurements which occur in nature often fit a Normal
Distribution – symmetrical, greatest frequency is in the middle.
E.g. height, weight, dimensions of the human body.
2 Parameters:
m: Mean
s: Standard Deviation
 Cannot calculate probabilities for specific values because the
distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. Year 13s in NZ whose height is between 170 and 180cm.
PROBABILITY
=
AREA UNDER THE CURVE
 We cannot calculate probabilities for specific values because
the distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. Year 13s in NZ whose height is between 170 and 180cm.
PROBABILITY
=
AREA UNDER THE CURVE
Standard Deviations, s
On a Normal Distribution Curve, about…
 ___% of the population lies within 1 standard deviation
either side of the mean, m.
X
m
 We cannot calculate probabilities for specific values because
the distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. Year 13s in NZ whose height is between 170 and 180cm.
PROBABILITY
=
AREA UNDER THE CURVE
Standard Deviations, s
On a Normal Distribution Curve, about…
 68% of the population lies within 1 standard deviation
either side of the mean, m.
 __% within 2 standard deviations of m.
X
m
 We cannot calculate probabilities for specific values because
the distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. Year 13s in NZ whose height is between 170 and 180cm.
PROBABILITY
=
AREA UNDER THE CURVE
Standard Deviations, s
On a Normal Distribution Curve, about…
 68% of the population lies within 1 standard deviation
either side of the mean, m.
 95% within 2 standard deviations of m.
 __% within 3 standard deviations of m.
X
m
 We cannot calculate probabilities for specific values because
the distribution is continuous – infinite degree of precision.
 We calculate probabilities for a domain of values.
E.g. Year 13s in NZ whose height is between 170 and 180cm.
PROBABILITY
=
AREA UNDER THE CURVE
Standard Deviations, s
On a Normal Distribution Curve, about…
 68% of the population lies within 1 standard deviation
either side of the mean, m.
 95% within 2 standard deviations of m.
 99% within 3 standard deviations of m.
X
m
The Standard Normal Distribution
Properties * Horizontal axis measures “z”, the number of standard
deviations away from the mean.
* Mean, m = 0
* Standard deviation, s = 1
* P(a < Z < b) = shaded area
a b
Use of Standard Normal Tables
- The tables give the value P(0 < Z < a).
- Diagrams are essential.
z = Number of Standard Deviations
from the mean, m.
0 a
Properties * Horizontal axis measures “z”, the number of standard
deviations away from the mean.
* Mean, m = 0
* Standard deviation, s = 1
* P(a < Z < b) = shaded area
a b
Use of Standard Normal Tables
- The tables give the value P(0 < Z < a).
-Diagrams are essential.
z = Number of Standard Deviations
from the mean, m.
Examples:
(a) P(0  Z  1) = ?
0 a
(c) P(0  Z  1) = ?
(c) P(0  Z  1) = 0.3413
Properties * Mean, m = 0
* Standard deviation, s = 1
* The curve is symmetrical
P(a < Z < b) = shaded area
a b
Use of Standard Normal Tables
- The tables give the value P(0 < Z < a).
-Diagrams are essential.
z : Number of Standard Deviations
from the mean, m.
Examples:
(a) P(0  Z  1) = 0.3413
0 a
(c) P(0  Z  1) = 0.3413
Properties * Mean, m = 0
* Standard deviation, s = 1
* The curve is symmetrical
P(a < Z < b) = shaded area
a b
Use of Standard Normal Tables
- The tables give the value P(0 < Z < a).
-Diagrams are essential.
z : Number of Standard Deviations
from the mean, m.
Examples:
(a) P(0  Z  1) = 0.3413
0 a
Use of Standard Normal Tables
- The tables give the value P(0 < Z < a).
-Diagrams are essential.
z : Number of Standard Deviations
from the mean, m.
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = ?
0 a
Use of Standard Normal Tables
- The tables give the value P(0 < Z < a).
-Diagrams are essential.
z : Number of Standard Deviations
from the mean, m.
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
0 a
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = ?
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0 <Z  3.2) – P(0<Z<0.3)
(c) P(0  Z  3.2) = ?
(c) P(0  Z  3.2) = 0.4993
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- ?
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- ?
(c) P(0  Z  0.3) = ?
(c) P(0  Z  0.3) = 0.1179
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- ?
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
Examples:
(a) P(0  Z  1) = 0.3413
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = ?
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215)
(c) P(-0.326  Z  0) = ?
(c) P(-0.326  Z  0) = 0.1255 + 0.0022
(c) P(-0.326  Z  0) = 0.1277
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215)
= 0.1277
+ ?
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215)
= 0.1277
+ ?
(c) P(0  Z  2.215) = ?
(c) P(0  Z  2.215) = 0.4864 + 0.0002
(c) P(0  Z  2.215) = 0.4866
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215)
= 0.1277
+ 0.4866
(b) P(-1  Z  1) = 2  0.3413
= 0.6826
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215)
= 0.1277
+ 0.4866
= 0.6143
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215)
= 0.1277
+ 0.4866
= 0.6143
(e) P(Z  0.342 ) = ?
(c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3)
= 0.4993
- 0.1179
= 0.3814
(d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215)
= 0.1277
+ 0.4866
= 0.6143
(e) P(Z > 0.342) = 0.5 – P(0 < Z < 0.342)
= 0.5 - ?
(e) P(Z > 0.342) = 0.5
= 0.5
=
– P(0 < Z < 0.342)
-
(c) P(0.3  Z  3.2) = 0.4993 – 0.1179
= 0.3814
HW:
1. In NuLake: Do pages 296 & 297 (intro to Normal
Distn).
(d)
P(-0.326  Z  2.215) = 0.1278 + 0.4866
= 0.6144
2. In Sigma – match-up
task: OLD edition - p98 (Ex.
7.1) – Q57 only.
Or NEW edition – p 440 (exercise A.01) – Q57
(e)only.
P(Z > 0.342) = 0.5 – P(0 < Z < 0.342)
= 0.5 - 0.1339
= 0.3661
Lesson 2: Solve normal distribution
problems
• Calculate probabilities using the normal
distribution – standardise and use tables.
Notes, NuLake pg. 300 Q30-37.
HW:
Finish NuLake Qs (30-37)
*Extension (once NuLake Q30-37 done): Sigma (new
edition): p358 – Ex. 17.01: Q4, 5, 10, 11, 14 & 15.
Calculating probabilities for ANY normallydistributed random variable X, with mean, m
Any normal random variable, X, can be
transformed into a standard normal
random variable by the formula:
Example 1
If X is normally distributed with μ = 8 and σ = 2, find the probability that
X takes a value greater than 11,
s=2
STANDARDISING IT:
s=1
P( X >11) = P( Z > 11 8 )
2
= P(Z > 1.5)
m=8
= 0.5 – P(0  Z  1.5)
mz=0
11
z=1.5
P(0  Z  1.5) = ?
P(0  Z  1.5) = 0.4332
Calculating probabilities for ANY normallydistributed random variable X, with mean, m
Any normal random variable, X, can be
transformed into a standard normal
random variable by the formula:
Example 1
If X is normally distributed with μ = 8 and σ = 2, find the probability that
X takes a value greater than 11,
STANDARDISING IT:
P( X >11) = P( Z >
11 8
2
s=2
)
s=1
= P(Z > 1.5)
= 0.5 – P(0  Z  1.5)
m=8
= 0.5 – 0.4332
mz=0
= 0.0668
answer
11
z=1.5
Example 1
If X is normally distributed with μ = 8 and σ = 2, find the probability that
X takes a value greater than 11,
STANDARDISING IT:
s=2
11 8 )
P( X >11) = P( Z >
2
= P(Z > 1.5)
= 0.5 – P(0  Z  1.5)
= 0.5 – 0.4332
= 0.0668
Example 2:
s=1
m=8
mz=0
11
z=1.5
answer
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P( X >11) = P( Z >
11 8
2
)
s=2
s=1
= P(Z > 1.5)
= 0.5 – P(0  Z  1.5)
m=8
= 0.5 – 0.4332
Example 2:
= 0.0668
11
mz=0
answer
z=1.5
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P(175  X  184) = P( 175  178  Z 
5
s=5
184  178 )
5
s=1
= P( -0.6  Z  1.2)
= P(-0.6 Z 0) + P( 0 Z  1.2)
175 m=178 184
0.6
mz=0
1.2
P(-0.6 Z 0) = ?
P(-0.6 Z 0) = 0.2258
STANDARDISING IT:
P( X >11) = P( Z >
11 8
2
)
s=2
s=1
= P(Z > 1.5)
= 0.5 – P(0  Z  1.5)
m=8
= 0.5 – 0.4332
Example 2:
= 0.0668
11
mz=0
answer
z=1.5
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P(175  X  184) = P( 175  178  Z 
5
s=5
184  178 )
5
s=1
= P( -0.6  Z  1.2)
= P(-0.6 Z 0) + P( 0 Z  1.2)
175 m=178 184
0.6
mz=0
1.2
STANDARDISING IT:
P( X >11) = P( Z >
11 8
2
)
s=2
s=1
= P(Z > 1.5)
= 0.5 – P(0  Z  1.5)
m=8
= 0.5 – 0.4332
Example 2:
= 0.0668
11
mz=0
answer
z=1.5
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P(175  X  184) = P( 175  178  Z 
5
s=5
184  178 )
5
s=1
= P( -0.6  Z  1.2)
= P(-0.6 Z 0) + P( 0 Z  1.2)
=
0.2258
+
?
175 m=178 184
0.6
mz=0
1.2
STANDARDISING IT:
P( X >11) = P( Z >
11 8
2
)
s=2
s=1
= P(Z > 1.5)
= 0.5 – P(0  Z  1.5)
m=8
= 0.5 – 0.4332
Example 2:
= 0.0668
11
mz=0
answer
z=1.5
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P(175  X  184) = P( 175  178  Z 
5
s=5
184  178 )
5
s=1
= P( -0.6  Z  1.2)
= P(-0.6 Z 0) + P( 0 Z  1.2)
=
0.2258
+
?
175 m=178 184
0.6
mz=0
1.2
P(0 Z 1.2) = ?
P(0 Z 1.2) = 0.3849
STANDARDISING IT:
P( X >11) = P( Z >
11 8
2
)
s=2
s=1
= P(Z > 1.5)
= 0.5 – P(0  Z  1.5)
= 0.5 – 0.4332
Example 2:
= 0.0668
answer
m=8
11
mz=0
z=1.5
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P(175  X  184) = P(
s=5
s=1
175  178
184  178
 Z
)
5
5
= P( -0.6  Z  1.2)
= P(-0.6 Z 0) + P( 0 Z  1.2)
= 0.2258 + 0.3849
175 m=178 184
0.6
mz=0
1.2
Example 2:
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P(175  X  184) = P(
s=5
175  178  Z  184  178 )
5
5
s=1
= P( -0.6  Z  1.2)
= P(-0.6 Z 0) + P( 0 Z  1.2)
= 0.2258
+
0.3849
175 m=178 184
0.6 mz=0
1.2
Example 2:
The heights of a class of Y13 males are normally distributed with a mean
of 178cm and standard deviation of 5cm. Find the probability that a
randomly picked student from the class is between 175cm & 184cm tall.
STANDARDISING IT:
P(175  X  184) = P(
s=5
175  178  Z  184  178 )
5
5
s=1
= P( -0.6  Z  1.2)
= P(-0.6 Z 0) + P( 0 Z  1.2)
= 0.2258
= 0.6107
+
0.3849
answer
175 m=178 184
0.6 mz=0
1.2
17.05A
Eg3: The weights of suitcases received by an airline at check-in can be
modelled by a normal distribution with mean 17 kg and standard
deviation 4 kg. The airline charges for excess baggage if a suitcase
is ‘overweight’—defined as weighing more than 20 kg.
Find the probability that two consecutive passengers both have to
pay for excess baggage.
Eg3: The weights of suitcases received by an airline at check-in can be
modelled by a normal distribution with mean 17 kg and standard
deviation 4 kg. The airline charges for excess baggage if a suitcase
is ‘overweight’—defined as weighing more than 20 kg.
Find the probability that two consecutive passengers both have to
pay for excess baggage.
P(excess) = P(X > 20)
= 0.2266
Eg3: The weights of suitcases received by an airline at check-in can be
modelled by a normal distribution with mean 17 kg and standard
deviation 4 kg. The airline charges for excess baggage if a suitcase
is ‘overweight’—defined as weighing more than 20 kg.
Find the probability that two consecutive passengers both have to
pay for excess baggage.
P(X > 20) = 0.2266
Draw a tree diagram to show the possibilities for two suitcases.
Eg3: The weights of suitcases received by an airline at check-in can be
Do Nulake pg. 300-303
modelled by a normal distribution with mean 17 kg and standard
Q3037 (MUST do).
deviation 4 kg. The airline charges for excess baggage if a suitcase
is ‘overweight’—defined as weighing more than 20 kg.
Extension: Sigma (new
Find the probability that two consecutive passengers both have to
edition): p358 – Ex. 17.01:
pay for excess baggage.
Q5, 10, 11, 14 & 15.
P(X > 20) = 0.2266
0.2266
0.7734
0.2266
Over weight
0.7734
Under weight
0.2266
Over weight
0.7734
Under weight
Over weight
*
Under weight
P(two cases overweight) = 0.2266  0.2266
= 0.0513
Using your Graphics Calc. for Standard
Normal problems
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
Using your Graphics Calc. for Standard
Normal problems
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
s=5
E.g. 1:
If m=178, s=5, P(175 X  184) = ?
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 m=178 184
Using your Graphics Calc. for Standard
Normal problems
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
s=5
E.g. 1:
If m=178, s=5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 m=178 184
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
s=5
E.g.1:
If m=178, s=5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 m=178
s=3.5
E.g.2:
If m=30, s=3.5, P(X  31) = ?
MENU, STAT, DIST, NORM, Ncd
lower: -EXP99, upper: 31, σ: 3.5, μ: 30
184
m=30 31
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
s=5
E.g.1:
If m=178, s=5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 m=178
s=3.5
E.g.2:
If m=30, s=3.5, P(X  31) = 0.61245
MENU, STAT, DIST, NORM, Ncd
lower: -EXP99, upper: 31, σ: 3.5, μ: 30
184
m=30 31
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
s=5
E.g.1:
If m=178, s=5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 m=178
s=3.5
E.g.2:
If m=30, s=3.5, P(X  31) = 0.61245
MENU, STAT, DIST, NORM, Ncd
lower: -EXP99, upper: 31, σ: 3.5, μ: 30
184
m=30 31