Mmm Solutions

7/27/2019 Mmm Solutions

1/16

1. A laboratory blood test is 95% effective in detecting a certain disease when it is, infact, present. However, the test also yields a false positive result for 1% of thehealthy people tested. If 0.5% of the population actually has the disease, what is theprobability a person has the disease given that the test was positive? Define the events:

D: the tested person has the disease E: the test result is positive

What do we want to compute? P(D|E)From the problem statement, what do we know about the probabilities?

P(E|D) = 0.95, P(E|Dc) = 0.01 P(D) = 0.005 P(Dc) = 0.995

Using the above, we can solve for the desired probability by applying Bayes Rule:

P(D|E) =P(E

|D)P(D)

P(E|D)P(D) + P(E|Dc)P(Dc)=

(0.95)(0.005)

(0.95)(0.005) + (0.01)(0.995) 0.323

So only 32% of those people whose test results are positive actually have the disease!

1


2/16

2. Suppose that of the potential new patients that call up a medical clinic, only 60%are able to immediately speak to someone. The other 40% leave a message with theirphone numbers, in which case 75% of the time their call is returned the same day,while 25% of the time it is returned the following day.

Data indicates that the probability that a caller will visit the clinic is 80% if their callwas picked up immediately, whereas it is 60% if the call was returned the same day,and 40% if it was returned the following day.

For the following problems, simplify it as much as possible but your final answer doesnot have to be a single number

(a) (20pts) What is the probability that a random caller will visit the clinic?

Define the events:

V: the event that a caller visits the clinic I: the event that a caller spoke to someone immediately S: the event that a caller had their call returned the same day F: the event that a caller had their call returned the following day

From the problem statement, we have that for those who spoke to someone im-mediately:

P(I) = 0.60 P(V|I) = 0.80and for those who did not speak to someone immediately:

P(S|Ic) = 0.75 P(F|Ic) = 0.25

andP(V|S Ic) = 0.60 P(V|F Ic) = 0.40

Using the definition of conditional probability we have that:

P(S Ic) = P(S|Ic)P(Ic) = (0.75)(0.40)

P(F Ic) = P(F|Ic)P(Ic) = (0.25)(0.40)Using the law of total probability we therefore have that:

P(V) = P(V|I)P(I) + P(V|S Ic)P(S Ic) + P(V|F Ic)P(F Ic)= (0.80)(0.60) + (0.60)(0.75)(0.40) + (0.40)(0.25)(0.40)= 0.70

2


3/16

(b) (5pts) Given that a caller visited the clinic, what is the probability that theyspoke to someone immediately when they called the clinic?

P(I

|V) =

P(V|I)P(I)P(V)

=(0.80)(0.60)

(0.80)(0.60) + (0.60)(0.75)(0.40) + (0.40)(0.25)(0.40)= 0.686

3


4/16

3. An airline finds that 4 percent of the passengers that make reservations on a particularflight will not show up. Consequently, their policy is to sell 100 reserved seats on aplane that has only 98 seats. Find the probability that every person who shows up forthe flight will find a seat available.

Let X count the number of passengers that show up. Then X is binomial (n =100, p = 0.96) random variable. The probability that everybody who shows up has aseat is:

P(X 98) = 1 P(X > 98)= 1 P(X = 99) P(X = 100)= 1

100

99

(0.96)99(0.04)1

100

100

(0.96)100(0.04)0

= 0.913

4


5/16

4. A buyer must decide whether or not to accept a large lot of items which, unknownto him, contains 5% defectives. The lot will be accepted if there are only 0 or 1defective items in a sample of 20 items randomly drawn from the lot. Calculate theexact probability that the lot will be accepted.

Let X count the number of defectives in the 20 items sampled. Define p = 0.05, thenX is a binomial(n = 20, p = 0.05) random variable, and the probability that the lot isaccepted is just:

P(X = 0) + P(X = 1) =

20

0

p0(1 p)20 +

20

1

p1(1 p)19

= P(X 1) = BINOMDIST(1, 20, 0.05, 1) = 0.736

5


6/16

5. Experience shows that 20% of the people reserving tables at a certain restaurant nevershow up. If the restaurant has 30 tables and takes 32 reservations, use the Bernoullitrials model to:

(a) determine the probability that the restaurant will be able to accomodate everyone.

We think of each person showing up or not as an independent Bernoulli trialwith probability of success p = 0.80. Since there are 32 reservations, we takethe number of trials to be n = 32. We can therefore model the number of peopleshowing up as a binomial random variable, X, with parameters (n = 32, p = 0.80).

The probability that the restaurant can accomodate everyone is thus:

P(X 30) = 1 P(X = 31) P(X = 32)= 1

32

31

(0.80)31(0.20)1

32

32

(0.80)32(0.20)0

(b) Find the expected number of reservations that will be fulfilled.

This is the mean of a binomial r.v. with parameters (n = 32, p = 0.80):

E[X] = np = 32 0.80 = 25.6

6


7/16

6. Suppose that the number of accidents occurring on a highway each day is a Poissonr.v with parameter = 3

(a) Find the probability that 3 or more accidents occur today (I want a number)

Let X count the number of accidents that occur. Then X is a Poisson r.v. withparameter = 3.

P(X 3) = 1 P(X < 3)= 1 P(X = 0) P(X = 1) P(X = 2)= 1 e33

0

0! e3 3

1

1! e3 3

2

2!= 0.576

(b) Repeat part(a) under the assumption that at least 1 accident occurs today (again,your answer should be a number)

P(X 3|X 1) = P(X 3, X 1)X 1

=P(X 3)P(X 1)

and:P(X 1) = 1 P(X = 0) = 1 e3 = 0.95

Thus: P(X 3|X 1) = 0.5760.95

= 0.607

7


8/16

7. The time (in minutes) until the next email arrives in your inbox is modeled as anexponential random variable. It is observed that the probability that it takes morethan 2 minutes for the next email to arrive is exactly 3

5the probability that it more

than 1 minute for it to arrive.

Given this information, find the probability that it takes more than 4 minutes for thenext email to arrive.Let X be the time until the next email arrives. We know that X is an exponentialrandom variable, but we do not know the parameter, , of the distribution. However,we can solve for it from the relationship:

3

5P(X > 1) = P(X > 2)

3

5(1 P(X 1)) = 1 P(X 2)

3

5e

= e2

3

5= e take ln of both sides

ln

3

5

=

which yields that = ln 35

. Now, we can solve the problem:

P(X > 4) = 1 P(X 4) = e4 = e4 ln( 35) =

3

5

4

8


9/16

8. (20pts) Suppose that the weight of a person selected at random from some populationis normally distributed with parameters and . Suppose also that P(X 160) = 0.5and P(X 140) = 0.25.

(a) (10pts) Find and

P(X 160) =

160

= 0.5

where () is the standard normal cdf. From Excel, we have that (0) = 0.5.Therefore:

160

= 0

or = 160. We also have that:

P(X

140) = 140

= 140 160

= 0.25Using Excel, we see that NORMSINV(0.25)=-0.67449. Therefore:

140 160

= 0.67449

We therefore obtain that = (140 160)/0.67449 = 29.65.(b) (5pts) Find P(X 200).

P(X 200) = 1P(X < 200) = 1200 16029.65 = 1(1.35) = 10.9115 = 0.0885

9


10/16

(c) (5pts) Of all the people in the population weighing at least 200 pounds, whatpercentage will weigh over 220 pounds?

P(X 220|X 200) = P(X 220, X 200)P(X 200) = P(X 220)P(X 200)=

1 22016029.65

0.0885

=1 (2.0236)

0.0885

1 0.97850.0885

= 0.243

10


11/16

9. (20pts) Suppose that the width of an item produced by a factory is normally distributedwith = 0.9000 and = 0.0030. The item is considered defective unless its width isin the specification limits 0.9000 0.0050.

(a) (10pts) What percentage of items produced will be defective?

Let X be the width of a randomly selected item. Then X N(, 2) with = 0.9000 and = 0.0030.

An item is defective ifX < 0.90000.0050 = 0.0050 or X > 0.9000+0.0050 = 0.0050. The probability that a random item is defective is thus:

P(X < 0.0050) + P(X > + 0.0050)= P(X < 0.0050) + 1 P(X + 0.0050)=

0.0050

+ 1

+ 0.0050

= 0.0050

+ 1 0.0050

= 2

1

0.0050

since (x) = 1 (x)

= 2

1

5

3

= 2(1 0.95221) = 0.095581

where the excel function NORMSDIST(53

) was used to evaluate the standardnormal cdf.

The percentage of defective items is just the probability that a randomly selecteditem is defective, so the answer is 9.5581 %.

(b) What is the maximum allowable value of that will permit no more than 1 in100 defectives when the widths are normally distributed with = 0.9000 and ?We want to determine the maximum value of such that this inequaility holds:

1

100= 0.01 2

1

0.0050

The larger , is the more likely the item will be defective ( is a measure of

variability, so increasing it will increase the probability of the width being largerthan the specification limits), so the maximum value of is the one that satisfies:

0.01 =2

1

0.0050

Rewrite this as:

0.0050

= 1 (0.01/2) = 0.995

11


12/16

then use the Excel function NORMSINV(0.995) which yields 2.5758. Set thisequal to 0.005/ and solve for to obtain that:

= 0.005/2.5758 = 0.001941

This is the maximum allowable standard deviation for the width of the itemsproduced if we want no more than 1% of them to be defective.

12


13/16

10. (20pts) The gross weekly sales at a certain restaurant is a normal random variablewith mean $2200 and standard deviation $230. Assume the gross sales in each weekare independent of the sales in any other week.

For this problem, it is not necessary to provide a numerical answer. However, to get

full credit, be sure to define any symbols, variables, or random variables you introducesuch that I could get a numerical answer with a calculator or computer within a fewseconds

(a) (10pts) What is the probability that the total gross sales over the next 2 weeksexceeds $5000?

The sum of 2 independent normal random variables is itself a normal randomvariable with mean 1 + 2 and variance

2

1+ 2

2. In this case 1 = 2 = $2200

and 1 = 2 = $230. Let X N(4400, 2(2302)) be the total gross sales for thenext 2 weeks. The probability that total gross sales for the next 2 weeks exceeds$5000 is then given by:

P(X > 5000) = 1 P(X 5000)= 1

5000 4400

2(230)

= 1 (1.844)= 0.0325

(b) (10pts) What is the probability that weekly sales exceed $2000 in at least 2 of the

next 3 weeks?Let Y N(2200, 2302) be the sales for a given week. Define p by:

p = P(Y > 2000) = 1 P(Y 2000) = 1

2000 2200230

= 0.808

Now let W count the number of times sales exceeds $2000 in the next 3 weeks:W is a binomial(n = 3, p) random variable. We want the probability that thatweekly sales exceed $2000 in at least 2 of the next 3 weeks. This is equivalent to:

P(W

2) = P(W = 2) + P(W = 3)

=

32

p2(1 p)1 +

33

p3(1 p)0

= 0.9033

13


14/16

11. (20pts) One has 100 light bulbs whose lifetimes are independent exponential randomvariables with mean 5 hours. If the bulbs are used one at a time, with a failed bulbbeing immediately replaced by a new one, what is the probability that there is still aworking bulb after 525 hours?

Note that if Xi is an exponential random variable with mean 5 hours, we know that = 1/5 = 0.20. We want to determine

P(X1 + X2 + X100 525)

that is, the probability that the sum of the lifetimes of all 100 bulbs is at least 525hours. In terms of the sample mean S100 =

100i=1 Xi/100, this is just:

P(100 S100 525) = P(S100 5.25)

Since the light bulbs are used independently, and assuming that n = 100 is largeenough, we can invoke the central limit theorem and approximate the probability withthe standard normal cdf:

P

Sn E[Sn]

Var(Sn) a

(a)

In our case, E[Sn] = 5 and Var(Sn) = Var(Xi)/n = 52/n = 25/n, where n = 100. We

then have that:

P(S100 5.25) =

5.25 525/100

= (0.50)

Therefore:P(X1 + X2 + X100 525) = 1 (0.5) = 0.3085

14


15/16

12. From past experience a professor knows that the test score of a student taking his finalexamination is a random variable with mean 75 and variance 25. Use the central limittheorem to determine how many students would have to take the examination so as toensure, with probability at least 0.90, that the class average would be within 5 of 75.

Let Sn =n

i=1 Xi/n. In our case E[Sn] = 75 and Var(Sn) = 25/n.

We use the Central Limit Approximation, such that:

P

Sn E[Sn]

Var(Sn) a

(a)

where Sn N(75, 25/n). The average of the class is defined to be Sn = (X1 + X2 + + Xn)/n, and we want to find n such that:

P(70 < Sn

80)

0.90

We can rewrite this as:

P(Sn 80) P(Sn 70) 0.90

In terms of the standard normal cdf, this is just:

P(Sn 80) P(Sn 70) =

80 7525/n

70 75

25/n

= (

n) (n) = 2(n) 1

We want to find n such that this is at least 0.90. To find the minimum n, set theprobability to 0.90:

2(

n) 1 = 0.90so

n =

1(0.95)2

= [NORMSINV(0.95)]2 = 2.7

so the professor needs at least 3 students to take the exam.

15


16/16

13. A change in the manufacturing process of steel beams is suggested. The manufacturerwants to know if the mean strength of the beams made by a new process is differentfrom the 42,000 psi, which is the mean strength of the beams made by the present pro-cess. A two-sided hypothesis test on is appropriate to answer this question. Breakingstrengths are measured for a test sample of n = 32 beams made by the new process.

The sample mean for this test sample is x = 42, 196 psi and the sample standard devi-ation is s = 500 psi. Assume that the observations are normally distributed and thatthe sample size is large enough such that s . At the 10% level of significance, testthe hypothesis that:

H0 : = 42, 000 vs H1 : = 42, 000

Proceed in three different ways:

(a) Calculate the z-statistic:

z =x

0

/n =42196

42000

500/32 = 2.217

For = 0.10, we compare this to the critical value z/2 = NORMSINV(0.95) =1.645. Since |z| > z0.05, we reject H0 and conclude that the mean breakingstrength of the beam made by the new process is different. (in fact, the testindicates it is larger).

(b) Calculate the p-value.

p = 2(1 (|z|)) = 2(1 NORMSDIST(|z|)) = 2(1 0.987) = 0.026

Since p < , reject H0.(c) Compute a 90% confidence interval:

x z/2 n

=

42, 196 1.645

500

32

= [42051, 42341]

Since 0 does not fall in the CI, we reject H0

16

Documents

Mmm Solutions