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7. Continuous Probability Distributions
Cumulative Distribution Function
We define a cumulative distribution function F^ ( x )of a random variable X as F ( x ) P ( X x ). Properties of the cdf
- F^ (^ )^0 , F (^ )^1
- F ( x )
Probability Density Function
If the cdf F^ ( x )is differentiable then we can define a probability density function f ( x ) by f ( x ) F '( x ) The density function satisfies the following conditions: f^ ( x ) is non-negative, The total area under the curve representing f^ ( x )equals 1.
The probability that X falls between a^ and b is found by calculating the area under
the graph of f^ ( x ) between a^ and b
b a P ( a X b ) f ( x ) dx The expected value of the random variable X is
x E X xf x dx all ( ) ( ) Properties of the expected value: E(c)=c E(aX)=aE(X) E(X+Y)=E(X)+E(Y) E(XY)=E(X)E(Y), if random variables X and Y are independent, i.e. P(X<a and Y<b)=P(X<a)P(X<b) for all possible a and b The expected value of the random variable g^ (^ X )is x EgX gxfx dx all (()) ()( ) In particular,
^
x x Var X E X EX x E X f x dx E X xf x dx all 2 2 all ( ) ( ) ( ) ( ) ( ) ( ) or 2 all all ( ) (^2 ) ( ( ))^22 ( ) ( )
x x Var X E X E X x f xdx xf x dx Properties of the variance: Var(const)= Var(aX)= a^2 Var(X) Var(X+Y)=Var(X)+Var(Y)+2COV(X,Y), where COV(X,Y)=E{(X-EX)(Y-EY)} If X and Y are independent, then COV(X,Y)=0 and Var(X+Y)=Var(X)+Var(Y) If Var(X)=0 , then X=const Uniform Distribution A random variable X is said to be uniformly distributed if its density function is The expected value and the variance of the uniform distribution is The Normal Distribution This is the most important continuous distribution considered here. Many random variables can be properly modeled as normally distributed. Many distributions can be approximated by a normal distribution. The normal distribution is the cornerstone distribution of statistical inference. Normal distribution
A random variable X with mean^ ^ and variance 2 is normally distributed if its
probability density function is given by
Normal random variable has a bell shaped distribution, symmetrical around^ ^.
. 1 ( ) a x b b a f x 2 12 (b a)^2 Var(X) a b E(X)
- 14159 ... 2. 71828 ...
2 2 2 ( )
where and e f x e x x
If X is normally distributed with mean ^ and the variance 2 , then aX^ is also
normally distributed with the mean a^ ^ and the variance a^2 ^2. If (^) X and (^) Y are normally distributed independent random variables with the means (^) X and (^) Y , and the variances 2 (^) X and 2 (^) Y , then aX bY also has the normal distribution with mean a^ ^ X ^ b^ Y and the variance 2 2 2 2 a (^) X b Y. Finding Normal Probabilities Two facts help calculate normal probabilities: The normal distribution is symmetrical. Any normal distribution can be transformed into a specific normal distribution called “STANDARD NORMAL DISTRIBUTION” or Z
Every normal variable with some^ ^ and^ ^ , can be transformed into the standard
normal random variable: By the properties mentioned above we have that E (^^ Z )^0 and Var^ (^ Z )^1. Therefore, once probabilities for (^) Z are calculated, probabilities of any normal variable can found. The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of the random variable Z.
Definition Let us consider a number^ ^ such that 0 1 , then critical value of standard
normal distribution z ^ represents that value for which the area under the standard
normal curve to the right of z ^ is equal to^ ^ , i.e.
P ( Z z ) Exponential Distribution The exponential distribution can be used to model
- the length of time between telephone calls
- the length of time between arrivals at a service station
- the life-time of electronic components When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution. X
Z X^ X
A random variable is exponentially distributed if its probability density function is given by f ( x ) e ^ x , x 0 where^ ^ is a parameter of the distribution, the average number of occurrences of the events E ( X )^1 Var ( X )^1 P ( X x ) e ^ ^ x Exercises: p. 240: 4.84—4.96; p. 241: 4.102—4.107. References:
1. Chase and Bown, General Statistics 2. Hildebrand and Ott, Statistical Thinking for Managers 3. Keller and Warrack, Statistics for Management and Economics 4. McClave, Benson, and Sincich, A First Course In Business Statistics Exercises
- The lengths X of the nails in a large shipment received by a carpenter are approximately normally distributed with mean of 2 inches and standard deviation .1 inch. (a) If a nail is randomly selected, find P (^^1.^8 ^ X ^2.^07 ). (b) What proportion of nails has lengths that lie within one standard deviation of the mean? (c) The carpenter cannot use a nail shorter than 1.75 inches or longer than 2.25 inches. What percentage of the shipment of nails will the carpenter be able to use?
- Scores of males on the 1974 Mathematical Scholastic Aptitude Test (MSAT) were normally distributed with mean 500 and standard deviation 100. (a) What score indicates a percentile rank of 95? (b) The middle 40% of the distribution is bounded by what two scores? (c) If 1000 of these students are randomly selected, how many are expected to score higher than 650?
- The length of time it takes for a ferry to reach a summer resort from the mainland is approximately normally distributed with mean 2 hours and standard deviation 12 minutes. Over many past trips, what proportion of times has the ferry reached the island in (a) Less than 1 hour, 45 minutes? (b) More than 2 hours, 5 minutes? (c) Between 1 hour, 50 minutes and 2 hours, 20 minutes?
- Some auto companies design emission sensors so that they must be replaced after 100,000 miles. One such company found that the service life X (in months) of these sensors is approximately randomly normal with mean 48 months and standard deviation 9 months. (a) The company decided to guarantee the sensors
