chapter 1 part 2 student
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STA408 Pn. Sanizah's Notes 3/9/2014
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TOPIC 2 PART 2
CONTINUOUS RANDOM
VARIABLES AND
CONTINUOUS PROBABILITY
DISTRIBUTIONS
Continuous random variable
• A random variable is called continuous if it can assume all possible values in the possible range of the random variable where the data can take infinitely many values.
• Examples of the continuous random variables are:
– The computer time (in seconds) required to process a certain program.
– The time for a baby to gain the weight of 5.5 kg.
– The amount of rain falls in a certain city.
– The amount of water passing through a pipe connected with a high level reservoir.
– The heat gained by a ceiling fan when it has worked for one hour. [email protected]
Probability density function• The probability function of the continuous random variable is
called the probability density function (p.d.f.).
• The number of possible outcomes of a continuous random variable is uncountable infinite (we calculate a probability for a continuous random variable over an interval and NOT for any particular point).
• This probability can be interpreted as an area under the graph between the interval from a to b.
• How to find the probability of some interval of the continuous random variable?
Use Integral calculus. [email protected] [email protected]
Probability Density Functionis given by the area of the shaded region.
The graph of f is the density curve.
( )y f x
ba
( )P a X b
4
Properties for a Continuous Probability Distribution
5)(
)(
)()(
bXaP
bXaP
bXaPbXaP
Note: For continuous probability distribution
Mean (expected value) and VARIANCE of a continuous r.v.
Let X be a continuous random variable with probability distribution f(x).
1. The mean or expected value of X is
2. The variance of X is
where
dxxfxXE )()(
222 )]([)()( XEXEXV
dxxfxXE )()( 22
STA408 Pn. Sanizah's Notes 3/9/2014
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The Normal DistributionX~N(, 2)
The normal distribution is the most widely known and used of all distributions. Since
the normal distribution approximates many natural phenomena so well, it has
developed into a standard of reference for many probability problems.
THE NORMAL DISTRIBUTION X~N(, 2)
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The normal distribution curve,
has the following features:
1. It is bell-shaped.
2. The curve is symmetric about the
mean, .
3. The highest point on the normal
curve is at the mean of the
distribution.
4. The standard deviation determines
the width of the curve.
5. The two tails of the curve extend
indefinitely from - to +.
6. The total area under the curve is
1.0.
mean variance
The Normal Distribution X~N(, 2)
• The random variable X has normal probability distribution if its probability density function is defined by:
9
2)(2
1
2
1)(
x
exf
Note constants:
=3.14159…
e=2.71828…
This is a bell shaped curve with
different centers and spreads
depending on the mean, and
standard deviation, .
- < x <
Areas Under the Normal Curve
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Each of the two shaded areas is 0.5 or 50%
0.5 0.5
x
The total shaded area is 1.0 or 100%
A normal curve is symmetric
about the mean.
b
adxxf )(Area
1)(Area
dxxf
5.0)(Area
dxxf 5.0)(Area
dxxf
Normal Distributions
11Bluman Chapter 6
THE NORMAL DISTRIBUTION X~N(, 2)
• The probability that a random variable X lies in the interval ato b is written as P(a < X < b). To find this probability, we need to find the area under the normal curve between a and b which is by integrating.
• To find probability for normal distribution:
PROBLEM: How to integrate f(x)???? DIFFICULT
Need to use STANDARD NORMAL TABLE12
0,,2
1xf
2
2
1
forewhere
dxxfbXaP
x
b
STA408 Pn. Sanizah's Notes 3/9/2014
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Normal DistributionX~N(, 2)
Standard Normal Distribution
Z~N(0, 1)
Standard Normal Distribution
• Since the normal probability function is complicated and very difficult to integrate, standard normal tables are used instead.
• Hence, the standard normal random variable, Z is defined by:
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)1,0(~where
deviation standard
meanvalue
NZX
Z
Z
The Standard Normal (Z) Distribution
Z~N(0, 1)
• Mean = 0; Standard deviation = 1• When x = , z = 0• Symmetric about z = 0• Values of z to the left of center are negative• Values of z to the right of center are positive• Total area under the curve is 1.
Now, let’s learn how to use the STANDARD NORMAL TABLE provided.
15-3 -2 -1 0 1 2 3
σ = 1
µ = 0
z
The Standard Normal Random Variable Z
with = 0 and = 1
Example 1 Draw sketches and use the standard normal table
provided to find the following probabilities:
a) P(Z ≥ 1.54) = 0.0618 (to the right of z = 1.54 )
b) P(Z -1.54) = 0.0618 (to the left of z = -1.54 )
c) P(Z < 0.65)
d) P(Z > -0.95)
e) P(1.22 Z 2.40)
f) P(0 Z 0.4)
g) P(-1.77 Z -0.88)
h) P(-0.6< Z < 0)
i) P(-0.75 Z 1.29)
j) P(-1.97 Z 0.86)[email protected]
Example 2
If the random variable X is normally distributed with mean 125 and variance 16, find the probabilities of
(a) P(X ≥ 130)
(b) P(X ≤ 128)
(c) P(120 ≤ X ≤ 128)
APPLICATIONS OF THE NORMAL DISTRIBUTION
STA408 Pn. Sanizah's Notes 3/9/2014
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Example 3
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The weights of packages of ground beef are normally
distributed with mean 1 pound and standard deviation
0.10. What is the probability that a randomly selected
package weighs between 0.80 and 0.85 pounds?
Solution:
X=weights of packages of ground beef
X~N(1, 0.102)
)85.080.0( XP
)5.12( ZP
0440.0
0228.00668.0
)2()5.1(
)25.1(
ZPZP
ZP
Using symmetry
Example 4
The life time of the batteries for a notebook computer under normal usage is normally distributed with mean 210 minutes and standard deviation 15 minutes.
a) What percentage of these batteries will have a
life time between 195 minutes and 240 minutes?
b) What percentage of batteries will have a life
time less than 180 minutes?
Example 5A local report stated that the mean score on a placement test was 480 and that 20% of the candidates scored below 400. Assume that the scores follow normal distribution.
a) Find the standard deviation of the scores.
b) Find the percentage of candidates that scored above 500.
A desktop PC used 120 watts of electricity per hour based on 4 hours of use per day the variable is approximately normally distributed and the standard deviation is 6. If 500 PCs are selected, approximately how many will use less than 106 watts of power
Example 6-8: Amount of Electricity Used by a PC
22Bluman Chapter 6
To qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores are normally distributed.
Example 6-9: Police Academy
23Bluman Chapter 6
HOMEWORK
• Redo Examples 6-1 – 6-5 using the table given in class.
• Exercises 6-1 Q41-46
• Redo Examples 6-6 and 6-7.