correct probability assignment

01 Jul correct probability assignment

1. MBB Problem 7.2 (modified)

The average (mean) amount of time that a manager at the Cimaron Valley Department of Human Services spends in the annual performance review with an employee is 27.5 minutes, with a standard deviation of 2.5 minutes (normal distribution).

(a) What percentage of the annual performance reviews in the department take between 25.0 and 30.0 minutes?

Solution;

Z= ( x- μ)/ σ, and probability is P(Z= ( x- μ)/σ)

μ=27.5 minutes, σ = 2.5 minutes

The percentage of annual performance reviews between 25.0 and 30.0 minutes can be calculated as ;

P(25.0<𝑋 > 30.0) = 𝑃 ( (25.0– 27.5) /2.5 )>Z> (30.0− 27.5) /2.5)

=P(-1>Z>1)= P(Z<1)- P(Z<-1)

= 0.84134-0.15866

= 0.68268 or 68.268%

The percentage of the annual performance reviews between 25.0 and 30.0 minutes is 68.268%

(b) Between 22.5 and 32.5 minutes?

Solution

P(22.5<𝑋 > 32.5) = 𝑃 ( (22.5– 27.5) /2.5 )>Z> (32.5− 27.5) /2.5)

=P(-2>Z>2)= P(Z<2)- P(Z<-2)

= 0.97725 – 0.02275

= 0.9545 or 95.45%

The percentage of the annual performance reviews between 22.5 and 32.50 minutes is 95.45%

(c) Between 20.0 and 35.0 minutes?

P(20.0<𝑋 > 35.0) = 𝑃 ( (20.0– 27.5) /2.5 )>Z> (35.0 − 27.5) /2.5)

=P(-3>Z>3)= P(Z<3) – P(Z<-3)

= 0.99865 – 0.00135

= 0.9973 or 99.73%

The percentage of the annual performance reviews between 22.5 and 32.50 minutes is 99.73%

(d) You can add two normal distributions together and the new distribution is also a normal distribution with a mean equal to the sum of the two means and a standard deviation equal to the square root of the sum of the two square roots (assuming that the two distributions are independent of each other). So, if the manager did two performance reviews in a row, the two reviews combined would last a mean of 27.5 + 27.5 = 55.0 minutes with a standard deviation of = 3.54 minutes, and if the manager did three reviews in a row, the three reviews combined would last a mean of 27.5 + 27.5 + 27.5 = 82.5 minutes with a standard deviation of = 4.33 minutes. If the manager schedules four performance reviews in a row every day starting at 10:00, what percentage of days would she expect to finish before 12:00.

Solution

For 4 performance reviews in a row

Mean = 27.5 + 27.5+27.5 + 27.5= 110 minutes

minutes

From 10:00 to 12:00, there are 2hours,( 120 minutes)

P(Z< (120− 110) /5) = P(Z<2) = 0.97725 or 97.725%

If the manager schedules four performance reviews in a row every day starting at 10:00, the percentage that would finish before 12:00 is 97.725%

2. MBB Problem 7.4 (modified)

Refer to Problem 7.3 (which we covered in lecture). The head of the Mariposa County Accounting Department wants to establish a standard regarding the length of time that employees can expect to wait to receive reimbursement for professional expenses.

(MBB Problem 7.3According to records kept by MariposaCounty, the average ( mean) amount of time that it takes for employees to be reimbursed for professional expenses incurred in service to the county is 36 days, with a standard deviation of 5 days. The distribution is normal. About 2 months ago, Latisha MacNeilattended a training conference for her job in the County Recreation Department. She filed for reimbursement of expenses that she incurred at the conference 42 days ago, but she has not received payment. What is the probability of receiving reimbursement within 42 days of filing?

 

(a) She wants to publish a standard that states the maximum number of days it takes the department to process 95% of the claims filed. Help her find that standard.

Solution

Z= ( x- μ)/ σ, and probability is P(Z= ( x- μ)/σ)

μ=36 days, σ = 5days

P(Z<( x- 36)/5)= 0.95

x=NORM.INV(0.95,36,5) = 44.22427 days

(b) Instead of the maximum number of days, she considers publishing a standard that states the range of days it takes the department to process 95% of the claims filed. Help her find that standard.

Solution

P((a- 36)/5 <Z<((b- 36)/5) = 0.95

-(a- 36)/5 <Z<((b- 36)/5) = 1.96

((b- 36)/5) <1.96, b=36+5*1.96 = 45.8days

-(a- 36)/5 <Z=1.96, a= 36- 5*1.96=26.2days

So, the range of days the department would take to process 95% of the claims filed is 26.2 days to 45.8days

(c) Which standard would be more useful to address employees’ concerns and why?

Solution

The standard that states the maximum number of days (44.22427 days) it takes process 95% of the claims .Taking a maximum of 44.22427 days to process simply implies that , claims can even processed as early the first day upto 44.22427 days , as compared to the standard where the range is 26.2 days to 45.8days. In other words , claims can be processed earlier under the standard which states maximum number of days.

3. MBB Problem 7.10 (modified)

The director of development at the Richman Children’s Science Museum has recently collected data on donations for the past several years. She finds that the data are normally distributed with a mean of $51 and a standard deviation of $14. The director’s ideal point for a minimum donation is $60.

(a) What percentage of individual donations are less than $60?

Solution

μ=$51, σ = $14

P(X<60)= P(Z< (60− 51)/14) = P(Z<(9/14)) = P(Z<0.642857142) = 0.73891 or 73.891%

(b) What percentage of individual donations are $60 or more?

Solution

P(60≤ x) = 1- P( X<60) = 1- 0.73891 = 0.26109 or 26.109%

(c) The director’s long-term goal for an average individual donation is at least $80. Based on the current percentage of donations that meet that goal, does achieving this goal seem reasonable in the immediate future?

Solution

P(80≤ x) = 1- P(X<80) = 1- P(Z< (80-51)/14) = 1- P( Z<2.07142871) = 1-0.98077= 0.01923 or 1.923%

This percentage is too low and so, it seems unreasonable of achieving the director’s long –term goal for an average individual donation of at least $80.

4. MBB Problem 8.4 (modified)

In a grand jury case, a bookstore was indicted in Oklahoma County on several counts of selling an obscene book. The grand jury was composed of 22 Baptists and 8 other people. The defendant feels that Baptists are biased against free speech.

(a) What is the probability that 22 or more Baptists are selected on a jury if Oklahoma County is 40% Baptist?

Solution

n=30,r=22 and probability of success=0.4

The jury case has is a binomial , where success is defined as being a Baptist.

Mean = np=30*0.4=12, variance=np(1-)= 30*0.6*0.4= 7.2,

Standard deviation = 2.683281573

we use normal approximation, because np is more than 10

P( X>or =22) =1-NORMDIST(21.5,12,2.683282,1)= 0.0002, ( using NORM.DIST excel

Function)

The probability of having 22or more Baptist selected is 0.0002

(b) Although the entire population is only 40% Baptist, the population available for jury duty is 60% Baptist. What is the probability that 22 or more Baptists are selected for the jury if the available jury duty population is 60% Baptist?

Solution

Probability of success = 0.6, variance = np*(1-p) =30*0.6*0.4= 7.2,

Standard deviation=2.683281573.

Mean = no= 30*0.6=18, which is more than 10, and therefore we use normal approximation

P( X>or =22)= 1-NORM.DIST(22,18,2.683281573,1)= 0.068019(using NORM.DIST excel

Function)

5. MBB Problem 8.16 (modified)

The Bluefield Regional Employment Service needs to place five individuals in jobs this week to meet its yearly quota. During the past several years, the service’s track record is that every person sent to interview for a job has a .6 probability of getting the job. The placements appear to be independent of each other.

a)The service decides to send seven individuals for interviews this week. Based on what you know, what is the probability that the service will make its yearly quota this week?

Solution

This is a binomial distribution, in which probability of success=0.6, n=7 , np=7*0.6= 4.2, variance = np(1-p) = 7*0.6*0.4 = 1.68, standard deviation= 1.29614814

The probability that the service will make its yearly quota this week can be computed as

P( 5 or more gets the job) = p(5)+P(6)+p(7)

P( 5)=C(7,5) *(0.6^5)*(0.4)^2=0.2612736

P(6)=C(7,6)*(0.6^6)*(0.4)^1 = 0.1306368

P(7)=C(7,7) *(0.6^7)*(0.4)^0 = 0.0279936

P(5 or more gets the job) = 0.2612736+0.1306368+0.0279936 = 0.419904

(b) The service would like to be at least 90% sure of meeting its quota. If it sends eleven individuals for interviews this week, can it be 90% sure?

Solution

P(5 or more gets job) =1-P( less than 5 gets job).

P( 4)=C(11,4) *(0.6^4)*(0.4)^7=0.070071091

P(3)=C(11,3)*(0.6^3)*(0.4)^8= 0.02335703

P(2)=C(11,2) *(0.6^2)*(0.4)^9 = 0.0051904512

P(1)=C(11,1) *(0.6^1)*(0.4)^10 = 0.00069206016

P(0)=C(11,0) *(0.6^ 0)*(0.4)^11 = 0.00004194304

1-{ P( 4)+ P(3)+ P(2)+ P(1)+ P(0)}= 0.900647424 or 90.0647424%

Yes , the service can be at least 90% sure of meeting its quota.

6. MBB Problem 9.10 (modified)

The mean number of contracts that consulting firms receive with the Department of Human Services is .5 for any given year.

a)Ecosystems Inc., a heavy contributor to political candidates, receives two contracts in one year. What is the probability that this event would occur by chance?

Solution

This is a poison distribution, where μ= 0.5(the average of successes),

In poison distribution, the probability is given by the formula

P(x; μ) = (e-μ) (μx) / x! , where μ= mean, x= number of success,

Plugging in the values ; P(2; 0.5) = (e-0.5) (0.52) / 2!  = 0.075816332 P(2; 0.5) = 0.0758(4 d.p)

(b) What is the probability that Ecosystems will receive four contracts in two years?

Solution

P(x; μ) = (e-μ) (μx) / x! ,

μ= 0.5 x= 4

Plugging in the values ; P(4; 0.5) = (e-0.5) (0.54) / 4!  = 0.001579506926 P(4; 0.5) = 0.00158 (4 d. p)

(c) What is the probability that Ecosystems will receive six contracts in three years?

Solution

P(x; μ) = (e-μ) (μx) / x! ,

μ= 0.5 ,x= 6

Plugging in the values ; P(4; 0.5) = (e-0.5) (0.56) / 6!  = 0.00002064299966

7. MBB Problem 9.16 (modified)

The University of Wisconsin employs 200 teaching assistants (TAs). If one-third of the TAs sign a petition calling for a collective bargaining election, an election will be held.

A.).A survey of 25 TAs indicates that 40% will sign the petition. What is the probability that a sample such as this could have occurred if one-third or fewer of the TAs in the population will sign such a petition?

Solution

This is hypergoemetric distribution, because the sample that we are drawing from is a finite number.N=200,k=200,/3, n=25

Mean= n*k/N= 25*(200/3)/200= 25/3

, 

Sd= sqrt.{25*200/3*(200-200/3)*175}/{200^2 *199}=4.8885538805

P( X< 1/3 or fewer ) = =NORM.DIST(80,200/3,4.8885538805,1)=0.996809

We use normal approximation

(b) The university surveys 25 additional TAs, so they now have a survey of 50 TAs, which also indicates that 40% will sign the petition. What is the probability that a sample such as this could have occurred if one-third or fewer of the TAs in the population will sign such a petition?

Solution

.N=200,k=200,/3, n=50

Mean= n*k/N= 50*(200/3)/200= 50/3

Sd= sqrt.{25*200/3*(200-200/3)*175}/{200^2 *199}=4.8885538805

P( X< 1/3 or fewer ) =NORM.DIST(160,200/3,4.8885538805,1)=1