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Type the characters you see in this image:Try different image Conditions of Use Privacy Policy © 1996-2014, Amazon.com, Inc. or its affiliates Let X repress the difference between the number of head and the number of tail optained .when a coin is tossed n time what are the possible values of X? For n=3 if the coin is assumed fair what are the probabilities associated with the values that X can take on. Section 3.1 - Random Variables[edit | edit source]Exercise 1[edit | edit source]Let be the number of beams that failed due to shear (S). Three beams are sampled, and they will fail either from shear (S) or from flexure (F). The sample space is: The random variable maps to the sample space as: Exercise 2[edit | edit source]Three examples of Bernoulli random variables: Exercise 3[edit | edit source]Consider an Experiment in which the number of pumps in use at each of two gas stations was determined. Define the random variables: Exercise 4[edit | edit source]Let be the number of nonzero digits in a randomly selected zip code. Exercise 5[edit | edit source]Consider a sample space for the number of rolls of a set of dice until the sum of 2 is rolled. Let . In this case the random variable for mapping the sample space to the real numbers is no infinite, it is discrete and dichotomous. Exercise 6[edit | edit source]is the number of cars observed at an intersection until a car turns Left (). (Cars can turn right (), left () or go straight ()). can take on the values 1, 2, 3, 4, ..., for example:
Exercise 7[edit | edit source]. Exercise 8[edit | edit source]Let be the total number of trails until you have three successes in a row. will take on the values Exercise 9[edit | edit source]Exercise 10[edit | edit source]The number of pumps in use at both a six-pump (Station 1) and a four-pump (Station 2) gas station will be examined.
Section 3.2 - Probability Distributions for Discrete Random Variables[edit | edit source]Exercise 11[edit | edit source]Let be the number of cylinders on the next car to be serviced.
Exercise 12[edit | edit source]Let be the number of ticketed passengers who show up for a flight.
The probability the third person on standby gets on the flight is: Exercise 13[edit | edit source]Let be the number of phone lines (out of six total lines) in use at any one time in the office. Using the given probability mass function:
Exercise 14[edit | edit source]Let be the number of forms the contractor needs to submit. Given the probability forms are required is proportional to via the function for .
Exercise 15[edit | edit source]Receive a shipment of 5 boards, select 2 boards for inspection at random.
The cumulative distribution function, cdf, is: Exercise 16[edit | edit source]Let be the number of homeowners who have insurance. The probability any one homeowner has insurance is 0.30.
Exercise 17[edit | edit source]90% of batteries have acceptable voltages. Let be the number of batteries tested to find two acceptable batteries.
Exercise 18[edit | edit source]Let be the maximum of two tosses of a fair six-sided die.
Exercise 19[edit | edit source]Let be the number of days after Wednesday it takes for both magazines to arrive. can take on the values 0, 1, 2, or 3. The probability mass function is: Exercise 20[edit | edit source]Three couples and two individuals are going to a seminar. Let be the number of people who arrive late for the seminar.
The CDF is:
Exercise 21[edit | edit source]Benford's Law,
The individual probabilities from the probability mass function are:
Exercise 22[edit | edit source]The CDF from Exercise 13 is: Exercise 23[edit | edit source]Using the CDF provided
Exercise 24[edit | edit source]
Exercise 25[edit | edit source]If a the boy is born first: If a the boy is born second then we have: If the boy is born third we have: If the boy is born fourth we have: The probability mass function for is: This is a preview of a distribution call the Geometric distribution. Exercise 26[edit | edit source]
Alvie will travel on straight line segments. He must travel segments to visit friends and then travel one more segment to get back home. Thus . The probability mass function for is:
If he visits a male first and then goes home, : probability of For visiting one female: Exercise 27[edit | edit source]Let the string (2, 4, 3, 1) denote the order that four textbooks are handed back to four students who claim to be missing their books. Let be the number of students who get their book back when the books are handed back in a random order.
Exercise 28[edit | edit source]Show that the cdf, , is a non-decreasing function. Let then: The equality holds if . Section 3.3 - Expected Values[edit | edit source]Exercise 29[edit | edit source]For the given probability mass function find the expectation and variance.
Exercise 30[edit | edit source]
Exercise 31.[edit | edit source]The expectation for the random variable in exercise 12 is: Find the probability is within one standard deviation of the mean. Exercise 32.[edit | edit source]For the pmf:
Exercise 33.[edit | edit source]Let be a Bernoulli random variable with success probability .
Exercise 34.[edit | edit source]For The expectation is: The expectation is finite. Exercise 35.[edit | edit source]For the pmf for the quantity of magazines is: If you make one dollar profit for each mag sold and you have to determine the number to order you have: The largest expected profit comes from purchasing 4 magazines to sell. Exercise 36.[edit | edit source]Expected damages incurred is: For an expected profit of $100 the premium amount charged should be $700. Exercise 37.[edit | edit source]Find the expectation and variance for the discrete uniform distribution. Exercise 38.[edit | edit source]The expected value of where is the roll of a fair die is: The guaranteed amount is So you are better off taking the gamble, the expected winnings are higher. Exercise 39.[edit | edit source]Let be the number of lots ordered. Let be the number of pounds left after lots are purchased. Exercise 40.[edit | edit source]The variance, a measure of spread, is the same for both and . Note that Exercise 41.[edit | edit source]Prove . Let . Exercise 42.[edit | edit source]Let and
Exercise 43.[edit | edit source]When then Exercise 44.[edit | edit source]Chebyshev's Inequality:
Exercise 45.[edit | edit source]If then show . Intuitively, we can consider the expectation to be the center of mass of the random variable. So if the mass of the random variable is between and , then the center of mass, the expectation, will also be between and . The proof is easy. You'll need to remember that Section 3.4 - The Binomial Probability Distribution[edit | edit source]Exercise 46.[edit | edit source]Find the following binomial probabilities.
Exercise 47.[edit | edit source]Find the following probabilities:
Exercise 48.[edit | edit source]Let be the number of defective compact disc players circuit boards in a random sample of size . The distribution of is:
Exercise 49.[edit | edit source]Let be the number of goblets which have flaws.
Exercise 50.[edit | edit source]Let be the number of calls involving a fax message.
Exercise 51.[edit | edit source]Let be the same as in Exercise 50.
Exercise 52.[edit | edit source]Let be the number of students who want a new textbook.
Exercise 53.[edit | edit source]Using the information in Exercise 30, the probability for any one person to be cited for moving violations while driving in the last three years is defined as: Find the following probabilities for a group of 15 such persons.
Exercise 54.[edit | edit source]Let be the number of people who want an oversize version of a tennis racket.
Exercise 55.[edit | edit source]Note: There are two solution methods presented here. The first is a little more complex than needed, but is a good way to illustrate several concepts in probability. We are told that 20% of all telephones of a certain type will be submitted for service under warranty. Of the telephones submitted for service 60% are repaired and 40% are replaced. If we by 10 telephones of this type, what is the probability that exactly 2 will be replaced under warranty? Start with defining some random variables. Let be the number of phones submitted for service. Let be the number of telephones which are replaced after being submitted for service. We have some conditional probabilities here and will need to use the law of total probability to find the final answer. We have the following distributions: Use the law of total probability to find the answer. Note that the first two probabilities in the sum are both zero, i.e., it is not possible to have if or .
Now that you've seen this solution, there is another way to do this that is a lot easier. Look at the probability that a telephone is replaced. Again, use the law of total probability. Now, let be the number of telephones replaced under warranty. and the probability that 2 telephones are replaced is: Exercise 56.[edit | edit source]Let be the number of students who receive special accommodations.
Exercise 57.[edit | edit source]We know that any one battery will have acceptable voltage with probability 0.90. A flashlight requires two batteries of acceptable voltage to work. If we assume that the batteries are independent the probability the batteries in a flashlight are acceptable is . Let be the number of working flashlights in a sample of size . The probability at least nine of the ten flashlights will work is: Exercise 58.[edit | edit source]Let be the number of defective components. Find the probability the batch is accepted given the different success probabilities . The sampling plan which maximizes the acceptance for and minimizes the acceptance for is the best option. Of the three methods suggested, it appears that method d, sampling 15 and accepting if less than 2 are defective is the best call. The ideal case would be having a step function with probability 1 of acceptance for and probability 0 of acceptance for . Exercise 59.[edit | edit source]This exercise is a prelude to hypothesis testing in the later chapters. In this case we have where is an unknown value. We will reject the claim of if we observe . .
Exercise 60.[edit | edit source]Let be the function for the total revenue from a total of passenger cars. The expected revenue of for 25 vehicles over the toll bridge is $40.00. Exercise 61.[edit | edit source]Let be the number of books received if the student writes on topic A. Let be the number of books received if the student writes on topic B. if then the probability of having at least half of the books ordered received is The probability of getting at least half of the ordered books for Topic B is slightly higher than for Topic A and thus Topic B should be selected. If the success probability for receiving a book is reduced from 0.90 to 0.50 the probabilities are: In this case the probability of getting at least half of the ordered books is higher for Topic A than B and Topic A should be selected. Exercise 62.[edit | edit source]Let be a binomial random variable. For a fixed value of is there a value of that will allow the variance to be zero? It turns out that there are two possible values for that will allow the variance to be zero. This should make sense. If the success probability is 0 then the out come of the experiment will be all failures, all the values are the same and the variance, a measure of spread, is zero since the spread of the data is zero. Same if the success probability is 1, all the outcomes are success, there is no spread in the data. To maximize the variance we need a value of . Here is the proof: Since the second derivative is less than zero for all values of we know that when the first derivative is zero that we will have a maximum. The maximum is when . Exercise 63.[edit | edit source]
The functions are the same. To do this define another random variable . The distributions are: Exercise 64.[edit | edit source]Prove the expectation of a binomial random variable is . Exercise 65.[edit | edit source]
Exercise 66.[edit | edit source]Let be the number of people with a reservation who show up for the limo.
Exercise 67.[edit | edit source]Using Chebyshev's Inequality find the lower bounds for and for The Chebyshev bounds are: For the binomial distribution we have: For the binomial distribution For the probability is 0.06523779 and for the probability is 0.003942142. Section 3.5 - Hyper geometric and Negative Binomial Distributions[edit | edit source]For all the ISE 261 Students ;) Section 3.6 - The Poisson Probability Distribution[edit | edit source]93a) Let P0(t,delta(t)) be the probability of no events (zero events) in the total time interval from 0 to t+delta(t). Let P0(t) be the probability of no events in the subinterval 0 to t. Let P0(delta(t)) be the probability of no events in the subinterval delta(t). Part I: what needs to occur in both subintervals is for no events to occur. Part II: P0(t) and P0(t,delta(t)) are not yet known so they are used as variables. but it is known that the probability of exactly 1 event occuring in the time frame delta(t) is alpha*delta(t). note: assume that delta(t) is so small that the occurrence of more than 1 event in delta(t) is zero i.e. o(delta(t)) is zero. so the probability of no events in the subinterval delta(t) is 1- alpha*delta(t). thus the total probability in the combined two subintervals of no events, assuming independence is P0(t,delta(t)) = P0(t)*(1-alpha*delta(t)). This the relationship. 93b) Expand and rearrange the relationship P0(t,delta(t)) = P0(t)*(1-alpha*delta(t)) to give (P0(t,delta(t)) - P0(t))/delta(t) = -alpha*P0(t) This defines the derivative of P0(t) as delta(t) goes to zero. 93c) it is easy to show that the derivative of P0(t) = e^(-alpha*t) is -alpha*P0(t). 93d) left side of the equation: differentiate given Pk(t) with respect to t using product rule. but first factor out constant alpha^(k)/k! then right side of equation: use factorial identity that (k-1)! = k!/k. its then just algebra and exponent math to show both sides are equal. Supplementary Exercises[edit | edit source] |