Statistics Assignment Essay Sample
To execute a certain type of blood analysis. lab technicians must execute two processs. The first process requires either one or two separate stairss. and the 2nd process requires either one. two. or three stairss. a. List the experimental results associated with executing the blood analysis. Answer: There are two processs that a lab technician must execute. The first process requires either one or two separate stairss. which could be named as x1 and x2. The 2nd process requires either one. two or three stairss. which could be named as y1. y2 and y3. The experimental results associated with executing the blood analysis from 1st process ( x1. x2 ) and 2nd process: ( y1. y2. y3 ) : * ( x1. y1 ) . x1. y2. x1. y3
* x2. y1. x2. y2. x2. y3
B. If the random variable of involvement is the entire figure of stairss required to make the complete analysis ( both processs ) . demo what value the random variable will presume for each of the experimental results. Answer: Valuess for the random variable to be assumed for each of the results. * ( x1. y1 ) =2 = & gt ; the 1st process required 1 measure and the second process required 1 measure * x1. y2=3 = & gt ; the 1st process required 1 measure and the second process required 2 stairss * x1. y3=4 = & gt ; the 1st process required 1 measure and the second process required 3 stairss * x2. y1=3 = & gt ; the 2nd process required 1 measure and the 1st process required 1 measure * x2. y2=4 = & gt ; the 2nd process required 1 measure and the 1st process required 2 stairss * x2. y3=5 = & gt ; the 2nd process required 1 measure and the 1st process required 3 stairss Chapter 5 – Section 2. Question 11
A technician services get offing machines at companies in the Phoenix country. Depending on the type of malfunction. the service call can take one. two. three. or four hours. The different types of malfunctions occur at about the same frequence. a. Develop a chance distribution for the continuance of a service call Answer: A chance distribution for the continuance of a service call. where ten is the continuance of the service call. and the chance of different malfunctions occur at the same frequence x| degree Fahrenheit ( x ) |
1| 0. 25|
2| 0. 25|
3| 0. 25|
4| 0. 25|
| 1. 00|
B. Pull a graph of the chance distribution.
Answer: A graph of the chance distribution
c. Show that your chance distribution satisfies the conditions required for a distinct chance map. Answer: Required conditions for a distinct chance maps are fx?0= & gt ; f1=f2=f3=f4=0. 25?0
?fx=1= & gt ; f1+f2+f3+f4=0. 25+0. 25+0. 25+0. 25=1
d. What is the chance that a service call will take three hours? Answer: The chance that a service call will take three hours is 0. 25 f3=0. 25
e. A service call has merely come in. but the type of malfunction is unknown. It is 3:00 autopsy and service technicians normally get away at 5:00 autopsy. What is the chance that the service technician will hold to work overtime to repair the machine today? Answer: The chance that the service technician will hold to work overtime to repair the machine today is 0. 5. Since it is 3pm right now and technician corsets overtime. so he might necessitate three or four hours to repair the machine. because if it takes merely one or two hr. the technician is done on clip. Therefore. the chance that it would take three or four hours: f3+f4=0. 25+0. 25=0. 5
Chapter 5 – Section 2. Question 13
A psychologist determined that the figure of Sessionss required to obtain the trust of a new patient is either 1. 2. or 3. Let ten be a random variable bespeaking the figure of Sessionss required to derive the patient’s trust. The undermentioned chance map has been proposed. fx=x6 for x=1. 2 or 3
a. Is this chance map valid? Explain.
Answer: This chance map is valid. because it meets the needed conditions for a distinct chance x?0= & gt ; f1=16?0
f2=26?0
f3=36?0
?fx=1= & gt ; f1+f2+f3=16+26+36=1
B. What is the chance that it takes precisely two Sessionss to derive the patient’s trust? Answer: The chance that is takes precisely two subdivisions to derive the patient’s trust is 0. 33. because Probability ( x=2 ) =f2=26=0. 3333
c. What is the chance that it takes at least two Sessionss to derive the patient’s trust? Answer: The chance that is takes at least two Sessionss to derive the patient’s trust is 0. 83. because Probability x?2=f2+f3=26+36=0. 8333=0. 83
Chapter 5 – Section 3. Question 17
a. Let ten be a random variable bespeaking the figure of times a pupil takes the SAT. Show the chance distribution for this random variable. Answer: The chance distribution:
Number of Times ( x ) | Number of Students| degree Fahrenheit ( x ) |
1| 721. 769| 0. 4752|
2| 601. 325| 0. 3959|
3| 166. 736| 0. 1098|
4| 22. 299| 0. 0147|
5| 6. 730| 0. 0044|
| 1. 518. 859| 1. 0000|
B. What is the chance that a pupil takes the SAT more than one clip? Answer: The chance that a pupil takes the SAT more so one clip is 0. 5248. because Probabilityx & gt ; 1=f2+f3+f4+f5=0. 3959+0. 1098+0. 0147+0. 0044=0. 5248 c. What is the chance that a pupil takes the SAT three or more times? Answer: The chance that a pupil takes the SAT three or more times is 0. 1289. because Probabilityx?3=f3+f4+f5=0. 1098+0. 0147+0. 0044=0. 1289
d. What is the expected value of the figure of times the SAT is taken? What is your reading of the expected value? Answer: The expected value of the figure of times the SAT is taken is 1. 6772. because Ex=?=?xfx
x| degree Fahrenheit ( x ) | xf ( x ) |
1| 0. 4752| 0. 4752|
2| 0. 3959| 0. 7918|
3| 0. 1098| 0. 3293|
4| 0. 0147| 0. 0587|
5| 0. 0044| 0. 0222|
| | 1. 6772|
Expected value or mean shows the norm of times a pupil takes the SAT. In this instance. the pupil is expected to take the SAT 1. 6772 times e. What is the discrepancy and standard divergence for the figure of times the SAT is taken? Answer: The discrepancy is 0. 5794 and standard divergence is 0. 7612 for the figure of times the SAT is taken. because Variance=Varx=o2=?x-?2fx
x| x-?| ( x-? ) 2| degree Fahrenheit ( x ) | x-?2f ( x ) |
1| -0. 6772| 0. 4586| 0. 4752| 0. 2179|
2| 0. 3228| 0. 1042| 0. 3959| 0. 0413|
3| 1. 3228| 1. 7497| 0. 1098| 0. 1921|
4| 2. 3228| 5. 3953| 0. 0147| 0. 0792|
5| 3. 3228| 11. 0408| 0. 0044| 0. 0489|
| | | | 0. 5794|
Standard Deviation= ?=Varx=0. 5794=0. 7612
Chapter 5 – Section 3. Question 23
a. What is the expected value of the figure of individuals populating in each type of unit? Answer: The expected value of the figure of individuals populating in Rent-Controlled is 1. 57 and in Rent-Stabilized is 2. 08 Number of Persons| Rent-Controlled| Rent-Stabilized|
| degree Fahrenheit ( x ) | xf ( x ) | xf ( x ) | xf ( x ) |
1| 0. 61| 0. 61| 0. 41| 0. 41|
2| 0. 27| 0. 54| 0. 30| 0. 6|
3| 0. 07| 0. 21| 0. 14| 0. 42|
4| 0. 04| 0. 16| 0. 11| 0. 44|
5| 0. 01| 0. 05| 0. 03| 0. 15|
6| 0. 00| 0| 0. 01| 0. 06|
| | 1. 57| | 2. 08|
B. What is the discrepancy of the figure of individuals populating in each type of unit? Answer: The discrepancy of the figure of individuals populating in Rent-Controlled is 0. 75 and in Rent-Stabilized is 1. 41 Number of Persons| Rent-Controlled|
x| x-?| ( x-? ) 2| degree Fahrenheit ( x ) | x-?2f ( x ) |
1| -0. 57| 0. 32| 0. 61| 0. 20|
2| 0. 43| 0. 18| 0. 27| 0. 05|
3| 1. 43| 2. 04| 0. 07| 0. 14|
4| 2. 43| 5. 90| 0. 04| 0. 24|
5| 3. 43| 11. 76| 0. 01| 0. 1176|
6| 4. 43| 19. 62| 0. 00| 0. 00|
| | | | 0. 75|
Number of Persons| Rent-Stabilized|
x| x-?| ( x-? ) 2| degree Fahrenheit ( x ) | x-?2f ( x ) |
1| -1. 08| 1. 17| 0. 41| 0. 48|
2| -0. 08| 0. 01| 0. 30| 0. 002|
3| 0. 92| 0. 85| 0. 14| 0. 12|
4| 1. 92| 3. 69| 0. 11| 0. 41|
5| 2. 92| 8. 53| 0. 03| 0. 26|
6| 3. 92| 15. 37| 0. 01| 0. 15|
| | | | 1. 41|
c. Make some comparings between the figure of individuals populating in rent-controlled units and the figure of individuals populating in rent-stabilized units. Answer: The mean figure of individuals populating in Rent-Stabilized lodging units is higher than figure of individuals populating in Rent-Controlled units. because 2. 08 & gt ; 1. 57. In footings variableness. the figure of individuals in Rent-Stabilized lodging units is besides higher. than in Rent-Controlled. 1. 41 & gt ; 0. 75 Chapter 5 – Section 4. Question 31
A Randstad/Harris synergistic study reported that 25 % of employees said their company is loyal to them ( USA Today. November 11. 2009 ) . Suppose 10 employees are selected indiscriminately and will be interviewed about company trueness. a. Is the choice of 10 employees a binomial experiment? Explain. Answer: The choice of 10 employees is a binomial experiment. because: 1. There is a sequence of 10 indistinguishable tests
2. Probability is the same for all 10 employees
3. Tests or employees are independent
B. What is the chance that none of the 10 employees will state their company is loyal to them? Answer: The chance that none of the 10 employees will state their company is loyal to them is 0. 0563. because fx=nxpx ( 1-p ) n-x
fx=0=1000. 250 ( 1-0. 25 ) 10-0=0. 0563
c. What is the chance that 4 of the 10 employees will state their company is loyal to them? Answer: The chance that 4 of the 10 employees will state their company is loyal to them is 0. 146. because fx=nxpx ( 1-p ) n-x
Px=4=1040. 254 ( 1-0. 25 ) 10-4=0. 145998001
d. What is the chance that at least 2 of the 10 employees will state their company is loyal to them? Answer: The chance that at least 2 of the 10 employees will state their company is loyal to them is 0. 756. because Px?2=P2+P3+P4+P5+P6+P7+P8+P9+P10
or
PX?2=1-PX & lt ; 2=1-P0-P1. where P0=0. 0563 from portion Angstrom
P1=1010. 251 ( 1-0. 25 ) 10-1=0. 187711715
so PX?2=1-PX & lt ; 2=1-P0-P1=1-0. 0563-0. 1877=0. 756
Chapter 5 – Section 3. Question 33
Twelve of the top 20 closers in the 2009 PGA Championship at Hazeltine National Golf Club in Chaska. Minnesota. used a Titleist trade name golf ball ( GolfBallTest website. November 12. 2009 ) . Suppose these consequences are representative of the chance that a indiscriminately selected PGA Tour participant uses a Titleist trade name golf ball. For a sample of 15 PGA Tour participants. do the undermentioned computations. a. Compute the chance that precisely 10 of the 15 PGA Tour participants use a Titleist trade name golf ball. Answer: The chance that precisely 10 of the 15 PGA Tour participants use a Titleist trade name golf ball is 0. 1859. because Px=nxpx ( 1-p ) n-x. where p=1220=0. 60
P0=15100. 6010 ( 1-0. 60 ) 15-10=0. 185937844
b. Compute the chance that more than 10 of the 15 PGA Tour participants use a Titleist trade name golf ball. Answer: The chance that more than 10 of the 15 PGA Tour participants use a Titleist trade name golf ball is 0. 2173. because Px & gt ; 10=P11+P12+P13+P14+P15
where
P11= 15110. 60111-0. 6015-11=0. 126775803
P12= 15120. 60121-0. 6015-12=0. 063387901
P13= 15130. 60131-0. 6015-13=0. 021941965
P14= 15140. 60141-0. 6015-14=0. 00470185
P15= 15150. 60151-0. 6015-15=0. 000470185
so
Px & gt ; 10=0. 126775803+0. 063387901+0. 021941965+0. 00470185+0. 0000470185=0. 217277704 c. For a sample of 15 PGA Tour participants. calculate the expected figure of participants who use a Titleist trade name golf ball. Answer: The expected figure of participants who use a Titleist trade name golf ball is 900. because Ex=np=15*0. 6=900
d. For a sample of 15 PGA Tour participants. calculate the discrepancy and standard divergence of the figure of participants who use a Titleist trade name golf ball. Answer: The discrepancy is 3. 6 and standard divergence is 1. 8974 of the figure of participants who use a Titleist trade name golf ball. because Varx=np1-p=15?0. 6?1-0. 6=3. 6
Standard Deviation=3. 6=1. 897366596
Chapter 5 – Section 5. Question 41
During the period of clip that a local university takes phone-in enrollments. calls come in at the rate of one every two proceedingss. a. What is the expected figure of calls in one hr?
Answer: The expected figure of calls in one hr is 30. because there are 60
proceedingss in one hr and calls come at the rate of every two proceedingss. so Ex=0. 5?60=30 calls per hr
B. What is the chance of three calls in five proceedingss?
Answer: Since the calls come in at the rate of one every two proceedingss. there would be 2. 5 calls in 5 proceedingss. because ?t=1?52=52=2. 5
so the chance of three calls in five proceedingss is 0. 2138. because Px=3=?ke-?x! =2. 53e-2. 53! =0. 213763017
c. What is the chance of no calls in a five-minute period? Answer: The chance of no calls in a five-minute period is 0. 0821 Px=0=?ke-?x! =2. 50e-2. 50! =0. 082084998
Chapter 5 – Section 5. Question 43
Airline riders arrive indiscriminately and independently at the passenger-screening installation at a major international airdrome. The average reaching rate is 10 riders per minute. a. Compute the chance of no reachings in a one-minute period Answer: The chance of no reachings in a one-minute period is 0. 0000454. because Px=0=?ke-?x! =100e-100! =0. 0000454
b. Compute the chance that three or fewer riders arrive in a one-minute period. Answer: The chance that three or fewer riders arrive in a one-minute period is 0. 0103. because
Px?3=P0+P1+P2+P3=0. 000453999+0. 002269996+0. 007566655=0. 01029065 where
P0=0. 0000454 from portion Angstrom
P1=?ke-?x! =101e-101! =0. 000453999
P2=102e-102! =0. 002269996
P ( 3 ) =103e-103! =0. 007566655
c. Compute the chance of no reachings in a 15-second period. Answer: The chance of no reachings in a 15-second period is 0. 0821. because P0=?ke-?x! =2. 50e-2. 50! =0. 082084998
?=15?1060=2. 5 arrivals*
*Since there are 60 seconds in one minute and I have a 15-second clip period d. Compute the chance of at least one reaching in a 15-second period. Answer: The chance of at least one reaching in a 15-second period is 0. 9179 Pno arrivals=1-P0=1-0. 082084998=0. 917915001
