Handout 4 (Chapter 4): Continuous Random Variables

STAT 211

Handout 4 (Chapter 4): Continuous Random Variables

A r.v. X is said to be continuous if its set of possible values is an entire interval of numbers. Then a probability distribution of X is f(x) (pdf: probability density function) such that for any two numbers a and b,

Conditions to be a legitimate probability distribution:

(i) f(x) ³ 0, for all x

(ii) (area under the entire graph of f(x)).

If X is a continuous r.v., then for any number c, P(X=c)=0.

Cumulative Distribution Function for X (cdf) : F(x)=

F(-¥)=0, F(¥)=1, , P(X>a)=P(X³a)=1-F(a)

Example: A college professor never finishes his lecture before the bell rings to end the period and always finishes his lectures within 2 min after the bell rings. Let X=the time that elapses between the bell and the end of the lecture and suppose the pdf of X is . Find the value of k which makes f(x) a legitimate pdf.

Example: The amount of bread (in hundreds of pounds) that a certain bakery is able to sell in a day is a random variable with probability function,

(a) Find the value of A which makes f(x) a legitimate pdf.

(b) What is the probability that the number of pounds of bread that will be sold tomorrow is

(i) more than 500 pounds

(ii) less than 500 pounds

(iii) between 250 and 750 pounds

Obtaining f(x) from F(x) : If X is a continuous r.v. with pdf f(x) and cdf F(x), then at every x at which the derivative exists, F`(x)=f(x).

Percentile of a continuous distribution: Let p be a number between 0 and 1. the (100p)th percentile of the distribution of a continuous r.v. X, denoted by r(p), is defined by

P(X£median)=P(X>median)=0.50

Expected value for the continuous random variable, X:

Variance for the continuous random variable, X:

or the shortcut is

If h(X) is the function of random variable X,

If h(X) is a linear function of X, the rules of the mean and the variance can directly be used instead of going through the mathematics.

Example: Go back to the example with the college professor in this chapter and calculate cumulative density function F(X), 90th percentile of X, the median, the expected value of X (E(X)) and the variance of X (Var(X)). Also demonstrate how to obtain f(x) from F(x). Define the findings by word.

Uniform Distribution: X ~U[a,b]

Chebyshev's Inequality : is the probability that the value of X lies at least k standard deviations from its mean is at most .

Empirical Rule: If the population distribution of a random variable is (approximately) normal, then

· Roughly 68% of the values are within one standard deviation of the mean

· Roughly 95% of the values are within two standard deviations of the mean

· Roughly 99.7% of the values are within three standard deviations of the mean

Normal Distribution: X ~ N (m , s2 )

A continuous r.v. X is said to have a normal distribution with parameters m and s2 where -¥ < m < ¥ and s > 0. The pdf of X is

It is symmetric and bell-shaped. It is called standard normal distribution when m=0 and s2 =1 and the random variable is denoted by Z (standard normal random variable,N(0,1)). The cdf of Z is =P(Z£z). Appendix table A.3 can be used to compute .

The (100p)th percentile of X with N (m ,s2) =

m + [The (100p)th percentile of Z with N (0,1)]×s

Example: Let X be normally distributed with mean, m=20 (symmetric around 20) and the variance, s2 =4 (or standard deviation, s=2), then is normally distributed with, m=0 and the variance, s2 =1 (or standard deviation, s=1).

This information can be shown as X~N(m=20, s2 =4) and Z~N(m=0, s2 =1)

(a) P(X<16) = P(X£16) because normal distribution is continuous.

P(X<16) = P(Z<-2)=0.0228

P(X>16) = 1 - P(X£16) = 1 - 0.0228 = 0.9772

(b) P(X<23.5) = P(Z<1.75) = 0.9599

P(X>23.5) = P(Z>1.75) = 1 - P(Z£1.75) = 1 - 0.9599 = 0.0401

P(X<x*)=0.05= then by looking at the z-table, =-1.645 Þ x* = 16.71. You need to subtract 0.71.

(d) If you would like largest 5% of X to be 24 how many points you need to add or subtract to X?

P(X>x*)=0.05= then by looking at the z-table, =1.645 Þ x* = 23.29. You need to add 0.71.

Example: In mathematics test, if we assume that your test scores (X) were approximately normally distributed with mean of 76.8 and standard deviation of 13.94.

a. If a score below 60 represents a grade of F (failure), approximately what percent of students failed the test?

b. If the cutoff for a grade of A is the lowest score of the top 15%, what is that cutoff point?

c. How many points must be added to the student scores so that only 10% fail (less than 60 be the failing grade)?

d. If the cutoff for a grade of C is the lowest score of the top 45%, what is that cutoff point?

e. What is the 90th percentile of test scores (x)?

Example:A chemical plant superintendent orders a process shutdown and setting readjustment whenever the pH of the final product falls below 6.9 or above 7.1. The sample pH is normally distributed with unknown m and standard deviation 0.05. Determine the probability

(a) of readjusting when the process is operating as intended and m=7

(b) of failing to readjust when the process is too alkaline and the mean pH is m=7.15

· IF X is a random variable whose logarithm is normally distributed then X has a lognormal distribution.

Normal Approximation to the Binomial Distribution:

Let X be a binomial r.v. based on n trials with success probability p. Then if the binomial probability histogram is not too skewed, X has approximately a normal distribution with m = n×p and then (check if np³10 and n(1-p)³10 to use the formula).

Exercise 4.48 (textbook): Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts and let X denote the number among these that are nonconforming and can be reworked. What is the approximate probability that X is

(a) At most 30?

(b) Less than 30?

The Gamma Distribution: X ~ Gamma ( a , b )

It is a skewed distribution. Note that

· If b=1 then it is called standard gamma distribution.

· When the random variable is a standard gamma r.v. then the cdf is called the incomplete gamma function (Appendix Table A.4). F(x;a,b)=F(x/b;a)

· If a = 1 then it is Exponential(b=1/l). It is used as a model for the distribution of times between the occurrence of successive events. The gamma distribution is used for the segment of time or and space occurring until some specified number of events has transpired where l being the average process rate and a being the specified number of events that must transpired as X is reached.

· If a = r / 2 and b = 2 then it is Chi-squared(r). (Appendix table A.7)

· If the random variable, X is distributed Exponential (b) then Y=X1/g has Weibull(g,b).

Exercise 4-56 (textbook): Suppose the time spent by a randomly selected student who uses a terminal connected to a local time sharing facility has a gamma distribution with mean 20 and variance 80 min2.

(a) What are the values of a and b?

(b) What is the probability that a student uses the terminal for at most 24 min?

Example: Find the probability that the time taken for the next two cars to arrive at a tollbooth will be 1 minute or less when l=2 per minute.

Example: Flaws in a reel of high-fidelity radar recording tape occur on the average of once every 10 feet. Determine the probability that the next recording will begin on a flawless stretch of tape over 5 feet long.

Example: A series system consists of 100 independent units, each with exponential distribution with l=0.005. Find the system reliability over a span of t=10.

Exercise 4-60 (textbook): Extensive experience with fans of a certain type used in diesel engines has suggested that the exponential distribution provides a good model for time until failure. Suppose the mean time until failure is 25000 hours. What is the probability that

(a) A randomly selected fan will last at least 20000 hours?

(b) A randomly selected fan will last at most 30000 hours?

(d) The lifetime of a fan exceeds the mean value by more than 2 standard deviations?

Exponential distribution shares memoryless property of the geometric distribution then P(X³t+t0|X³t0)=P(X³t). It simply is the distribution of additional lifetime is exactly the same as the original distribution of lifetime.

Probability plots: Order the n-sample observations from smallest to largest. Then the ith smallest observation in the list is taken to be the [100(i-0.5)/n]th sample percentile.

Exercise 4-80 (textbook): Ten observations on bearing lifetime (in hours) are collected. Construct a normal probability plot and comment on the plausibility of the normal distribution as a model for bearing lifetime.

Data / i / (i-0.5)/n / z
152.7 / 1 / .05 / -1.645
172 / 2 / .15 / -1.036
172.5 / 3 / .25 / -0.675
173.3 / 4 / .35 / -0.385
193 / 5 / .45 / -0.126
204.7 / 6 / .55 / 0.126
216.5 / 7 / .65 / 0.385
243.9 / 8 / .75 / 0.675
262.6 / 9 / .85 / 1.036
422.6 / 10 / .95 / 1.645

Exercise 4-87 (textbook): The failure time observations (1000’s of hours) resulted from accelerated life testing of 16 integrated circuit chips of a certain type. Use the corresponding percentiles of the exponential distribution with l=1 to construct a probability plot. Comment on the sample having been generated from any exponential distribution.

F(x)=

That means if the smallest 5 th percentile is observed then F(x)=0.05 and we are trying to find what x is. x can be found as -ln(1-F(x)).

Data / i / (i-0.5)/n / x
11.6 / 1 / 5/160 / 0.031749
26.5 / 2 / 15/160 / 0.09844
82.8 / 3 / 25/160 / 0.169899
179.7 / 4 / 35/160 / 0.24686
204.6 / 5 / 45/160 / 0.330242
212.6 / 6 / 55/160 / 0.421213
229.9 / 7 / 65/160 / 0.521297
242 / 8 / 75/160 / 0.632523
244.8 / 9 / 85/160 / 0.757686
304.3 / 10 / 95/160 / 0.900787
307.8 / 11 / 105/160 / 1.067841
359.5 / 12 / 115/160 / 1.268511
366.7 / 13 / 125/160 / 1.519826
379.1 / 14 / 135/160 / 1.856298
502.5 / 15 / 145/160 / 2.367124
558.9 / 16 / 155/160 / 3.465736