How to use binomial and poisson Tables: Tables are cumulative. Imagine each one (k variable) as a tile and you are laying tile on a floor. If k=2 than we have to
subtract tile k=1 (which s the cumulative length of tile 0 and 1) from k=2 so we know how big the tile is by itself. Now if the k=L is 0.287 that is the end of that tile. So
when we say k=2 is 0544 that is the end of the accumulated tile length of tile 0 and 1. So if the
Use this example which is based on poisson:
If A (also called y or probability)
P(x<1)=0.287 (or X=1(ork=1)
)
P(x > 1)= 1-0.082= 0.918 [So 1- (x=0)]
Questions from Test 2:
1. Ithas been estimated that 20% of patients develops an infection after being operated on at the local hospital. Since the number of infections is assumed to be
abinomial random variable, calculate the following probabilities for the next 10 operations. n=20 p=0.20 X ~ Binomial
a. P(0 infections) = 0.107
b. P(3 or fewer infections) = 0.879
c. P(4 or more infections) = 1 - P(X< 3) = 1- 0879 = 0.121
d. If you actually found 4 infectionsin the next 10 operations, what would you suggest we tell the hospital administrator? Since P(X>4) =0.121, it
appears that the number of infections (4)is consistent with the estimated 20% infection rate.
2. As people age they develop spots on their skin. For people between the ages of 60 and 70 it has been estimated that the average number of spots developed
on a normal person is 5.5 . Furthermore, it has been determined that the number of spots on a normal person follows a poisson distribution. If X is the number
of spots on a randomly selected person between the ages of 60 and 70, calculate the following probabilities. Sample Unit = “normal person” , X = # Spots on
normal person , X ~ poisson , A =5.5
a. P(X< 6) =0.686
b. P(X = 6) = P(X < 6) - P(X <5) = 0.686 - 0.529 = 0.157
c. P(X>3)=1-P(X <2)=1-0.088 =0.912
d. ‘The average number of spots for people who have a skin disease is much larger than the average for normal people. Therefore, doctors want to
use the number of spots on someone to help assess whether or not the patient has a skin disease. How many spots would a patient need to have before you would
suggest that they are not a normal person and may have a skin disease? Explain your answer. For normal people (A = 5.5 spots/person). Therefore since P(X <
11) = 0.989 we would expect normal people to have 11 or fewer spots approximately 99% of the time. If a person had more than 11 spots, we could
assume that they may be at risk for a skin disease.
Normal Distrioution ou conduct sailing trips for couples (one husband and one wife). Unfortunately, the Texas Parks and Wildlife has a 1000 pound
ur passengers (not counting yourself). The weight of husbands is known to have a mean of 190 Ibs and standard deviation of 20
weight of 130 Ibs and standard deviation of 10 lbs. If you sign up 3 couples (3 husbands and 3 wives), answer the following
Saowenr /| he weighis of husbands and wives are independent and nommally distributed.
1ge total weight of the 3 couples on the sailboat?
H1+H2+H3+W1+W2+W3
800 50 900 950 1000 1050 1100 +190 + 130 + 130 + 130 =960 Ibs
T
. Whatis the variance of the total weight
of the 3 couples?
1= 400 + 400 + 400 + 100 + 100 + 100= 1500
C. ‘What
is the standard deviation of the total weight of the 3 couples?
or= SQRT(1500) =38.7
d. If the game warden showed up (with scales to weigh your passengers) right when you were getting ready to go sailing with your 3 couples, what
is the probability that your 3 couples EXCEED the maximum weight limit of 1000 Ibs? Draw a picture ofthe total weight of the 3 couples to help you calculate
this probability.
T~Normal pr=960 , or=38.7 , X = 1000
Z.= (1000 — 960) / 38.7
P(X >1000) =P(Z
> 1.03) =1-0.8485 = 0.1515
15.15% probability that your boat is overloaded.
4. 1f arandom variable X has a normal distribution with mean 50 and std dev 2, calculate the following probabilities.
a. P(X < 53)= 0.9332
Z=(53-50)/2= L5 (z-score...looking for probability so look at z-table where row 1.5 is at column 1)
b. P(X < 47)= 0.0668
Z=(47-50)/2=-15
c. P(X = 50) = 0 (The chance of a number being exactly the mean (50.0000000000) is 0)
d. P(X > 60) =~ 0. really= (0.000000287)
Normal Distribution Z=(60-50)/2=5
€. P(6<X <52)=0.8185
6-50)/2= -2, which is 0.0228
2-50)/2=1, which s 0.8413.
0.8413-0.0228= 0.8185 (space between the two)
Normal Distribution 5. The amount of active ingredient in a blood pressure pill is known to be approximately normally distributed with mean 30
mg, and standard deviation 0.5 mg.
. Drawa picture ofthis normal distribution [be sure to label the X-axis and show the + 30 limits].
n+30=30+3%05=3L5
Pixs 309 ~ 1.0 p-30=30-3*0.5=285 Picture:
//' Target
2 275 25 285 @9 295 0 (08) 31
, b. Doctors discover that patients develop a rash if the blood pressure pill has more than 30.5 mg. of active ingredient. The probability that the pill
you take today has more than 30.5 mg, and you develop a rash can be shown to be approximately 16% ? With so many patients developing rashes from taking
these pills, the company puts you in charge of fixing this problem. After forming a process improvement team, the first suggestion s to change the process and
farget a mean less than the current 30 mg. What would you suggest for a new target mean such that almost ll the pills have less than 30.5 mg? [remember that
‘more than 30.5 mg will cause a rash] Draw a picture to show what you are recommending here.
Jir=Target Mean
6:=0.5 (assume std dev does not change)
pr=30.5-3%0; =305~ 1.5=29.0 Picture:
Normal Distrbution 6. Doctors at the UT Medical Research University have determined that the time to complete a specific operation
. is normally distributed with mean 100 minutes and std deviation of 5 minutes. Doctors who complete this operation in much
\ Less time may be taking shortcuts and putting the patient at risk. It was decided to measure the time it takes each doctor to do
| ppe— this operation for their next 25 patients. Plans are to estimate the average time [X-bar] it takes each doctor to complete the
operation on these 25 patiens.
a. Whats the sampling distribution of the sample mean? Be complete in your answer.
OEIACK - Population Parameters:
Sampling Distribution of x
b. Draw a picture of this sampling distribution and label the 3 sigma limits.
Pibar 3% G= 100
£ 3*1 = 100 +3 =97, 103] - >
c. 1f Doctor Wisecracker takes an average [X-bar] of 105 minutes to do the operation on his 25 patients, what would you suggest is going on.
Z=(105-100)/1=5/1=5
Since P(X-bar > 105) ~ 0, Wisecracker appears to take a lot longer than the typical doctors. In fact it appears that Wisecracker takes an average of 5 minutes
longer than the typical doctor.
d. If the time to complete this operation was not normally distributed, would this change your answers above? Explain your answer
If populationis not normally distributed, the sampling distribution of X-bar most likely would NOT be normally distributed because the central limit might not
apply for the small sample size of 25, Therefore can't work this problemin this course.
Module 14 — Problems — Linear Regression
Data was collected on students (50) GPA in the required math courses and their grade (numerical) made in statistics. As a professor, I am interested in building a model
that would allow me to predict a students grade in statistics as a function of their math GPA. A linear regression model was fit using Excel resulting in the output below.
a. Whats the model estimated?
Bo+Bi*X.
52.79139146 + 7.88856273(X)
T wrote this out completely so you can see them in the data below....should put in significant digits so § =52.791+ 7.889(X).
b. If your GPA in required math courses was 2.7, predict your grade in statistics.
L §=Bo+Bi*X
=52.791 +7.889 (2.7).
791 +21.3003
c. Is a students GPAin math statistically significant with respect to predicting their gradein statistics? Explain your answer.
‘We set this up to test the hypothesis h;: B; =0. We want this to be proved wrong! We want the slope (p;) to not be ‘0’. (‘0’ means flat...we don’t want that) So we want
to look at the P-value of the Intercept. WHAT?- go to the line/row that says ‘X Varaible 1’ go over to the P-value column and that numberis .0000042418 or 42418 E-
06. (This could also be written as 4.2418/6. All mean the same thing.) That is less than .05!
d. How much of the variabilityin statistics grades has been explained by this model?
‘What
is R*? In this question it is
SUMMARY OUTPUT
Regression
Statistics
Multiple R 0.598465505
R Square 0.358160961
Adjusted R Square 0.344789314
Standard Error 3.459534284
Observations 50
ANOVA
df SS MS F Significance F
Regression 1 320.5742483 320.5742483 26.7851051 4.42418E-06
Residual 48 574.4821183 11.96837746
Total 49 895.0563666
Coefficients Standard Error t Stat. P-value
Intercept 52.79139146 4.288796079 12.30914002 1.84708E-16
X Variable 1 7.88856273 1.524232459 5.175432842 4.42418E-06
subtract tile k=1 (which s the cumulative length of tile 0 and 1) from k=2 so we know how big the tile is by itself. Now if the k=L is 0.287 that is the end of that tile. So
when we say k=2 is 0544 that is the end of the accumulated tile length of tile 0 and 1. So if the
Use this example which is based on poisson:
If A (also called y or probability)
P(x<1)=0.287 (or X=1(ork=1)
)
P(x > 1)= 1-0.082= 0.918 [So 1- (x=0)]
Questions from Test 2:
1. Ithas been estimated that 20% of patients develops an infection after being operated on at the local hospital. Since the number of infections is assumed to be
abinomial random variable, calculate the following probabilities for the next 10 operations. n=20 p=0.20 X ~ Binomial
a. P(0 infections) = 0.107
b. P(3 or fewer infections) = 0.879
c. P(4 or more infections) = 1 - P(X< 3) = 1- 0879 = 0.121
d. If you actually found 4 infectionsin the next 10 operations, what would you suggest we tell the hospital administrator? Since P(X>4) =0.121, it
appears that the number of infections (4)is consistent with the estimated 20% infection rate.
2. As people age they develop spots on their skin. For people between the ages of 60 and 70 it has been estimated that the average number of spots developed
on a normal person is 5.5 . Furthermore, it has been determined that the number of spots on a normal person follows a poisson distribution. If X is the number
of spots on a randomly selected person between the ages of 60 and 70, calculate the following probabilities. Sample Unit = “normal person” , X = # Spots on
normal person , X ~ poisson , A =5.5
a. P(X< 6) =0.686
b. P(X = 6) = P(X < 6) - P(X <5) = 0.686 - 0.529 = 0.157
c. P(X>3)=1-P(X <2)=1-0.088 =0.912
d. ‘The average number of spots for people who have a skin disease is much larger than the average for normal people. Therefore, doctors want to
use the number of spots on someone to help assess whether or not the patient has a skin disease. How many spots would a patient need to have before you would
suggest that they are not a normal person and may have a skin disease? Explain your answer. For normal people (A = 5.5 spots/person). Therefore since P(X <
11) = 0.989 we would expect normal people to have 11 or fewer spots approximately 99% of the time. If a person had more than 11 spots, we could
assume that they may be at risk for a skin disease.
Normal Distrioution ou conduct sailing trips for couples (one husband and one wife). Unfortunately, the Texas Parks and Wildlife has a 1000 pound
ur passengers (not counting yourself). The weight of husbands is known to have a mean of 190 Ibs and standard deviation of 20
weight of 130 Ibs and standard deviation of 10 lbs. If you sign up 3 couples (3 husbands and 3 wives), answer the following
Saowenr /| he weighis of husbands and wives are independent and nommally distributed.
1ge total weight of the 3 couples on the sailboat?
H1+H2+H3+W1+W2+W3
800 50 900 950 1000 1050 1100 +190 + 130 + 130 + 130 =960 Ibs
T
. Whatis the variance of the total weight
of the 3 couples?
1= 400 + 400 + 400 + 100 + 100 + 100= 1500
C. ‘What
is the standard deviation of the total weight of the 3 couples?
or= SQRT(1500) =38.7
d. If the game warden showed up (with scales to weigh your passengers) right when you were getting ready to go sailing with your 3 couples, what
is the probability that your 3 couples EXCEED the maximum weight limit of 1000 Ibs? Draw a picture ofthe total weight of the 3 couples to help you calculate
this probability.
T~Normal pr=960 , or=38.7 , X = 1000
Z.= (1000 — 960) / 38.7
P(X >1000) =P(Z
> 1.03) =1-0.8485 = 0.1515
15.15% probability that your boat is overloaded.
4. 1f arandom variable X has a normal distribution with mean 50 and std dev 2, calculate the following probabilities.
a. P(X < 53)= 0.9332
Z=(53-50)/2= L5 (z-score...looking for probability so look at z-table where row 1.5 is at column 1)
b. P(X < 47)= 0.0668
Z=(47-50)/2=-15
c. P(X = 50) = 0 (The chance of a number being exactly the mean (50.0000000000) is 0)
d. P(X > 60) =~ 0. really= (0.000000287)
Normal Distribution Z=(60-50)/2=5
€. P(6<X <52)=0.8185
6-50)/2= -2, which is 0.0228
2-50)/2=1, which s 0.8413.
0.8413-0.0228= 0.8185 (space between the two)
Normal Distribution 5. The amount of active ingredient in a blood pressure pill is known to be approximately normally distributed with mean 30
mg, and standard deviation 0.5 mg.
. Drawa picture ofthis normal distribution [be sure to label the X-axis and show the + 30 limits].
n+30=30+3%05=3L5
Pixs 309 ~ 1.0 p-30=30-3*0.5=285 Picture:
//' Target
2 275 25 285 @9 295 0 (08) 31
, b. Doctors discover that patients develop a rash if the blood pressure pill has more than 30.5 mg. of active ingredient. The probability that the pill
you take today has more than 30.5 mg, and you develop a rash can be shown to be approximately 16% ? With so many patients developing rashes from taking
these pills, the company puts you in charge of fixing this problem. After forming a process improvement team, the first suggestion s to change the process and
farget a mean less than the current 30 mg. What would you suggest for a new target mean such that almost ll the pills have less than 30.5 mg? [remember that
‘more than 30.5 mg will cause a rash] Draw a picture to show what you are recommending here.
Jir=Target Mean
6:=0.5 (assume std dev does not change)
pr=30.5-3%0; =305~ 1.5=29.0 Picture:
Normal Distrbution 6. Doctors at the UT Medical Research University have determined that the time to complete a specific operation
. is normally distributed with mean 100 minutes and std deviation of 5 minutes. Doctors who complete this operation in much
\ Less time may be taking shortcuts and putting the patient at risk. It was decided to measure the time it takes each doctor to do
| ppe— this operation for their next 25 patients. Plans are to estimate the average time [X-bar] it takes each doctor to complete the
operation on these 25 patiens.
a. Whats the sampling distribution of the sample mean? Be complete in your answer.
OEIACK - Population Parameters:
Sampling Distribution of x
b. Draw a picture of this sampling distribution and label the 3 sigma limits.
Pibar 3% G= 100
£ 3*1 = 100 +3 =97, 103] - >
c. 1f Doctor Wisecracker takes an average [X-bar] of 105 minutes to do the operation on his 25 patients, what would you suggest is going on.
Z=(105-100)/1=5/1=5
Since P(X-bar > 105) ~ 0, Wisecracker appears to take a lot longer than the typical doctors. In fact it appears that Wisecracker takes an average of 5 minutes
longer than the typical doctor.
d. If the time to complete this operation was not normally distributed, would this change your answers above? Explain your answer
If populationis not normally distributed, the sampling distribution of X-bar most likely would NOT be normally distributed because the central limit might not
apply for the small sample size of 25, Therefore can't work this problemin this course.
Module 14 — Problems — Linear Regression
Data was collected on students (50) GPA in the required math courses and their grade (numerical) made in statistics. As a professor, I am interested in building a model
that would allow me to predict a students grade in statistics as a function of their math GPA. A linear regression model was fit using Excel resulting in the output below.
a. Whats the model estimated?
Bo+Bi*X.
52.79139146 + 7.88856273(X)
T wrote this out completely so you can see them in the data below....should put in significant digits so § =52.791+ 7.889(X).
b. If your GPA in required math courses was 2.7, predict your grade in statistics.
L §=Bo+Bi*X
=52.791 +7.889 (2.7).
791 +21.3003
c. Is a students GPAin math statistically significant with respect to predicting their gradein statistics? Explain your answer.
‘We set this up to test the hypothesis h;: B; =0. We want this to be proved wrong! We want the slope (p;) to not be ‘0’. (‘0’ means flat...we don’t want that) So we want
to look at the P-value of the Intercept. WHAT?- go to the line/row that says ‘X Varaible 1’ go over to the P-value column and that numberis .0000042418 or 42418 E-
06. (This could also be written as 4.2418/6. All mean the same thing.) That is less than .05!
d. How much of the variabilityin statistics grades has been explained by this model?
‘What
is R*? In this question it is
SUMMARY OUTPUT
Regression
Statistics
Multiple R 0.598465505
R Square 0.358160961
Adjusted R Square 0.344789314
Standard Error 3.459534284
Observations 50
ANOVA
df SS MS F Significance F
Regression 1 320.5742483 320.5742483 26.7851051 4.42418E-06
Residual 48 574.4821183 11.96837746
Total 49 895.0563666
Coefficients Standard Error t Stat. P-value
Intercept 52.79139146 4.288796079 12.30914002 1.84708E-16
X Variable 1 7.88856273 1.524232459 5.175432842 4.42418E-06
