Some Problems

Pulak Ghosh-IIMB

In a random sample of the length (in mins) of songs on CDs produced by hard rock bands, mean is 3.35 and s.d is 0.5.

Find the approximate proporHon of observaHons between 2.35 and 4.35 Ans: at least 1-(1/4)=3/4 of the observaHons Approximately what proporHon of songs last more than 5 mins. Ans: at most 1-0.91=0.09 (or 9%) of the observaHons are outside of this interval, either less than 1.7 or greater than 5. Since we can not assume that distribuHon is symmetric, we do not know what part of the 9% is less than 1.7 and what part is more than 5. To be conservaHve, the best we can say is at most 9% of the observaHons are more than 5 mins long. HINT: Use Chebyshevs empirical rule

MarkeHng research by Coee Beanery in Detroit, Michigan, indicates 70% of all customers put sugar in their coee, 35% add milk, and 25% use both. Suppose a Coee Beanery customer is selected at random.

What is the probability the customer uses at least one of the items

Ans:0.80 What is the probability the customer uses neither Ans: 0.20 What is the probability the customer uses just sugar Ans: 0.45 What is the probability the customer uses just one of these

two items Ans: 0.55

Over the past 15 years, there has been a decline in tradiHonal benets packages oered to workers (for example, lump sum payments upon reHrement) and an increase in the number of dened contribuHon plans. According to the Department of Labor, 43% of all workers in the private sector have a dened contribuHon plan and 14% have both dened-contribuHon plan and dened benets package. Suppose a worker in the private sector is selected at random. If the worker has a dened-contribuHon plan, what is the probability the worker also has a dened-benets plan?

Ans: Use condiHonal probability it is 0.325

Suppose that 18% of employees of a given corporaHon engage in physical exercise during the lunch hour. Moreover, assume that 57% of all employees are male, and 12% of all employees are males who engage in physical exercise during the lunch hour.

If we choose an employee at random from this corporaHon then what is the probability that this person is a female who engages in physical exercise during the lunch hour?

If we choose an employee at random from this corporaHon then what is the probability that this person is a female who does not engage in physical exercise during the lunch hour?

What proporHon of female employees engage in physical exercise during the lunch hour?

What proporHon of employees who engage in physical exercise during the lunch hour are male?

HINT:n see next slide!!!

Provided probabiliHes in the table ..

Exercise No Exercise All Male 12% 45% 57% Female 6% 37% 43% All 18% 82% 100% One issue with this quesHon is how probability quesHons can be worded. One might ask: What proporHon of employees are female? The answer is 43%. I could also ask if I choose an employee at random, what is the probability that they are female? The answer is again 43%. (a) The proporHon of employees who are female and exercise is 6%. (b) The proporHon of employees who are female and do not exercise during the lunch hour is 37%? (c) This quesHon restricts ahenHon to females only. This is a condiHonal probability. Out of the 43% of female employees, 6% exercise so the (condiHonal) probability is 6/43=14%. In terms of a formula (d) This quesHon restricts ahenHon to 18% of employees who exercise. The proporHon of these who are male is 12% so the (condiHonal) probability of being male is 67%. In terms of a formula

The probability a randomly selected family belongs to the AAA automobile club is 0.25. If a family belongs to AAA, the probability they have more than one car is 0.45. Suppose a family is randomly selected. What is the probability they have more than one car and belong to AAA?

Ans: Use joint probability (not condiHonal!!) It is 0.1125

Beher Bedding in East Hariord, ConnecHcut, claims 99.4% of all mahress deliveries are on Hme. Suppose two mahress deliveries are selected at random.

What is the probability both mahresses will be delivered on Hme

Ans: 0.994*0.994=0.988 What is the probability both mahresses will be delivered late Ans: 0.006*0.006=0.000036 What is the probability exactly one mahress will be delivered

on Hme Ans: 0.0119

A picnic is arranged to be held on a parHcular day. The weather forecast suggests that there is a 70% chance of rain on that day. If it rains, probability of good picnic is 0.2, while if it does not rain, probability of is 0.9

What is the probability that picnic will be good Ans: 0.41 If the picnic was good, what is the probability that it did not

rain on that day Ans: 0.65

Suppose 20% of clerical sta in an oce smoke cigarrehes. Research shows that 60% of smokers and 15% of nonsmokers suer a breathing illness by age 65.

(a) Do these percentages indicate that smoking and breathing illness are independent?

(b) what's the probability that a randomly selected 65 year old employee who has breathing illness smokes.

In a manufacturing plant, there are three machines producing 50%, 30% and 20% respecHvely of the total output. Out of the items produced by 1st machine, 4% are defecHve. Corresponding percentages for 2nd and 3rd machines are 5 and 3 respecHvely. An item is drawn at random from the producHon line.

Find the probability that this item is defecHve . Given that this item defecHve, what is the condiHonal

probability that it has been produced by machine 2?

You are trying to assess the eecHveness of an adverHsing campaign for a product which is bought by 20% of people. A survey shows that 40% of people recognise your adverHsement while 12% of people both recognise your adverHsement and buy your product.

Does seeing the adverHsement increase the probability of purchasing?

For another adverHsement, 30% of people recognise it

while 10% both recognise it and buy the product. Which of the two ads seems to be the most eecHve?

A manufacturing process produces computer chips in large numbers. Over the long-run the fracHon of bad chips produced by the process is around 20%. Thorough tesHng is expensive but there is a cheap test. All good chips pass the cheap test but so do 10% of bad chips.

Given that a chip passes the test what is the probability that it is a good chip?

If a company sold all chips that passed the test what fracHon of those they sold would be bad?

Hartneh car company has 12% market share. They are trying to assess the value of their long-standing adverHsing campaign. A standard way of doing this is to ask purchasers whether they recall the adverHsing campaign at the point of purchase. It turns out that 75% can recall the main details of the campaign. They also have general market research data which indicates that amongst those people who do not own a Hartneh, only 21% recall the major details of the campaign. On average, Hartneh makes a prot of k$12,200 on each customer.

What proporHon of people overall recall the major details of the campaign?

Given that someone recalls the major details of the campaign, what is the probability that they will buy a Hartneh?

On average, Hartneh makes a prot of k$12,200 on 12% of the car buying populaHon so their average prot per member of this target populaHon is 12% of k$12,200 which is $1,460. Given that someone recalls the major details of the campaign, what is the average prot per member of the target populaHon?

Set up a Bayes table. The two states of the world are that someone is a buyer or a non-buyer. The data is that the major details of the adverHsing campaign are recalled.

State Prior DL Product Posterior Value Buyer 0.12 0.75 0.090 0.328 $12.2 Non-Buyer 0.88 0.21 0.185 0.672 $0.0 $1.464 0.275 1.000 $3.996 The total of the product column gives the probability of the data here the

probability that someone recalls the campaign is 27.5%.

The posterior distribuHon gives the probability of the states given the data. CondiHonal on recalling the campaign, the probability of being a buyer is 32.8% - much higher than the overrall 12%.

Just calculate 32.8% of k$12,200 which is $3,996. The average value of a target customer to the company increases by $2,532 if they can just recall the details of the campaign.

Bayes in Market Research You are considering launching a new product- male asershave fragrance. It is very

dicult for new products in this market. On the basis of your experience, expert judgment and (to some extent) historical records, you guesHmate that there is around 10% chance your product will be a success. It is much more likely to be a op. To try and reduce your risk, you are considering retaining a markeHng company to do some test markeHng based on product tesHng and focus groups.

The company claims the following accuracy rates: 99% of ulHmately successful products will test well and return and overall posiHve recommendaHon. On the other hand, 89% of products that ulHmately op will test poorly and return an overall negaHve recommendaHon.

Q-1: How much does the companys service help you to reduce the

uncertainty in the products outcome? Does it actually reduce uncertainty at all?

Q-2: How much would the markeHng companys service be worth to you? For the sake of argument, suppose that a success makes you INR 1,000,000 and a op loses you INR 100,00

Example: Disease detecHon

17

A game is being played between two players A and B according to following rule: Each of them rolls a die with faces 1,2,..,6. If the total score is a perfect square, then A receives as many rupees as total score, otherwise B receives from A as many rupees as total score.

Find the probability distribuHon of the gain of A Find the mean gain for A

The Hard Rock Caf in Dallas carefully monitors customers orders and has found that 70% of all customers ask for some kind of coee (C) while the remainder order a specialized tea (T ). Suppose four customers are selected at random. Let the random variable X be the number of customers who order coee

Find the probability distribuHon of X Find the probability more than two customers order coee Suppose at least two customers order coee. What is the

probability all four customers order coee?

A red and a black die are thrown, and X denote the sum of the two dice. What is the Xs expected value, variance and standard deviaHon? What fracHon of the probability mass lies

within one standard deviaHon of the expected value?

Suppose the discrete random variable X, the age of a randomly selected child at the UniKids Academy Day Care Center has the probability distribuHon given in the following:

x: 1 2 3 4 5 6 7 f(x): 0.05 0.10 0.15 0.25 0.20 0.15 0.10 Find the expected value and variance of X Find the probability the random variable X takes on a value within one standard deviaHon of the mean

An insurance salesman meets with 5 prospecHve customers each week. From historical data, the proporHon of these customers who take out a policy is 18%. What is the probability that the salesman will sell 3 or more policies in a

given week? What is the probability that the salesman will sell 12 or more policies in a

given month (assuming a month is exactly 4 weeks)? What is the mean and standard deviaHon of the number of policies he

sells in a week, a month, a year? What do you noHce about these three distribuHons?

Using the binomial distribuHon with n=8 and p=0.18, the probability of 3 or more is 0.161.

In 4 weeks the number of trials (i.e. customers) is 32. Using the binomial distribuHon with n=32 and p=0.18, the probability of 12 or more is 0.007.

In 52 weeks the number of trials (i.e. customers) is 416. Using the binomial distribuHon with n=416 and p=0.18, the probability of 156 or more is zero to 8 signicant gures. While it is not at all impossible to make three sales in a week (see part (a)), it is virtually impossible for him to average 3 sales per week (by selling 156 in 52 weeks).

Over a week the mean is 1.4 and the standard deviaHon is 1.1. Over a month, the mean is 5.8 and the standard deviaHon is 2.17. Over a year, the mean is 74.9 and the standard deviaHon is 7.84. Obviously the mean number of sales increases in proporHon to the number of weeks. But the standard deviaHon does not. The standard deviaHon of 7.84 over a year is a much smaller proporHon of the mean of 74.8 during a year, than it is for a week. The yearly sales are much more predictable than the weekly sales.

Can a Binomially distributed r.v. have expectaHon 10 and variance 3?

Three contractors A, B, and C are bidding for n contracts. Suppose A has exactly half the chance as B; B has 4/5th as likely chance as C to win a contract. Results for dierent contracts are independent

What is the probability that A will win at least one contract Find the expected dierence in numbers of contracts to be

won by A and B

Researchers conducHng a study of buying habits interviewed thousands of customers and claim that 60% of all supermarket purchases are completely unplanned, or impulse purchase. Suppose 100 customers are selected at random

Find the mean, variance and std deviaHon of the number of customers who make an impulse purchase

Suppose 55 of the 100 customers make an impulse purchase. Is there any evidence to suggest the studys claim is false?

You survey 500 customers on whether or not they recognize your brand. On the basis of historical norms you expect 35% or 175 to recognize your brand. But you have been spending more money than usual on adverHsing. How many more than 175 would start to convince you that the extra money had been worth the expense? Is it feasible?

The number who recognize your brand is binomial with n=500 and, if nothing has changed, p=0.35. The mean is 175 and the standard deviaHon is 10.7. You would start to think things have improved if your results are say two standard deviaHons above the mean say around 175+22=197. The chance of 197 or more is 0.023 which is unlikely enough that you would think the campaign has worked. Even with 190 you would be prehy convinced since the chance of doing this well by luck is 0.088.

XYZ mutual fund has outperformed the index in 37 of the past 52 weeks. On this basis they are claiming that they can systemaHcally outperform the market. What do you think of this claim? Assuming that XYZ are not outperforming the market, what is

the chance they would get to make this claim during a 10 year period?

XYZ mutual fund has outperformed the index in 37 of the past 52 weeks? Think of each week as a trial. You either outperform or you do not. Each week is independent because nancial performances in each week are independent (a theorem of nance). What is the chance XYZ outperform the market in a given week? If they really are systemaHc outperformers then the probability would be high. But if they are just guessing then the chance of outperforming will not be high. Let us assume that there is a 50% chance of someone outperforming the market by sheer luck. We can vary this slightly if we need to. With these assumpHons, the number of weeks where XYZ outperform the market is binomial with n=52 and p=0.5.

The chance of 37 or more success is 1-binomdist(36,52,.5,true) which is 0.0182 which is prehy low. If we change the chance of outperforming the market by luck to 555 then this changes to 0.085 which is not so low.

In each year they get to make the claim or not. The years are now trials and the chance of success is 0.0182. The chance of at least one successful year in 5 equals 1 minus the chance of zero successful years. The answer is 0.088 from typing 1-binomdist(0,5,.0182,true). With 55% chance of outperforming this is 0.359.