Assignment Week3 - DS

\(Homework 3\)

1. mean steer weight=1300lbs variance of the steer weight=40,000 Therefore, standard deviation of steer weights= \(\sqrt(40,000) = 200lbs\) boundary weight = 979lbs z=(979-1300)/200=-1.605 P(Z>-1.605)=1-P(Z<-1.605)

pnorm(-1.605)

## [1] 0.0542469

P(Z<-1.605)=0.0543 Therefore, P(Z>-1.605)=0.9458

2. Mean life span = 11,000

Variance = 1,960,000,

Therefore, the standard deviation of the life span = sqrt(1960000)=1400

Boundary value=8340

Z=(8340 - 11000)/1400=>-1.9

P(Z>-1.9) = 1-P(Z<-1.9)

=>1-0.065615815=>0.9344

3. Mean income of firms= 80m, standard dev. Of incomes = 3m

Boundary values=83m & 85m

Therefore, z1=(83-80)/3=>1, z2=(85-80)/3=>1.67

P(z1 0.95254 - 0.841345=> 0.1112

4. score mean=456, standard deviation of 123

P(Z>z)=0.14=1-P(ZP(Z<z)=0.86

z=P-1 (0.86)=> 1.080319

let x= the minimum marks required to make it to the top 14%

z=(x-mean)/SD=>

(x-456)/123=1.080319

Therefore, x=589

5. mean length = 6.13 cms, standard deviation of lengths = 0.06 cms

P(Z>z1) = 1-0.07=>0.93, P(Z<z2)=0.07

z1=P-1 (0.93)=> 1.475791

z2= P-1 (0.07)=> -1.475791

if L1 & L2 are the lengths that correspond to standard variable Z’s values z1 & z2 respectively, then

L1=6.22cms, L2=6.04cms

6. mean scores=78.8, standard deviation of scores =9.8

Grade C: Scores below the top 45% and above the bottom 20%

P(Z>z1)=0.45=>1-P(Z<z1)=0.45=>P(Z<z1)=0.55

Therefore, z1= 0.125661

P(Z<z2)=0.20

Therefore, z2= -0.84162

if M1 & M2 are the marks that correspond to standard variable Z’s values z1 & z2 respectively, then

M1=80, M2=71

7. Mean score of 21.2, standard deviation of score = 5.4

P(Z>z)=0.45=1-P(ZP(Z<z)=0.55

z=P-1 (0.55)=> 0.125661

let x= the minimum marks required to make it to the top 45%

z=(x-mean)/SD=>

(x-21.2)/5.4= 0.125661

Therefore, x=21.9

8. Since this is a normal distribution function, we can use the binomial distribution with parameters ‘size’ and ‘probability’ as follows:

This is conventionally interpreted as the number of ‘successes’ in size trials.

Total number of trials=151, probability of success=0.09, number of successes=10

Binomial(10, 151, 0.09) = 0.192

9. since we are taking a sample of 147 tires, we would need the standard error of the mean.

mean lifetime of a tire is 48 months with a standard deviation of 7

SE of mean= sqrt(variance/N) => SD / sqrt(N).

SE of mean=7/sqrt(147)=>0.5774

So now the task is to obtain the probability of getting values>48.83, given mean=48 and SD’=0.5774

P(X>48.83) = 1-P(X<48.83)=>1-0.924709=0.075291

10. Since we are taking a sample of 68 computers, we would need the standard error of the mean.

Mean lifetime of a computer is 91 months with a standard deviation of 10 months

SE of mean= sqrt(variance/N) => SD / sqrt(N).

SE of mean=10/sqrt(68)=>1.2126

So now the task is to obtain the probability of getting values>93.54, given mean=91 and SD’=1.2126

P(X>93.54) = 1-P(X<93.54)=>1- 0.9819= 0.0181

11. Applying Fisher transformation using the below expression:

=> 1/2*(log((1+r)/(1-r))

Standard error:

=>1/sqrt(n-3)

Lets find the values for 4% and 10%

x10=1/2*log[(1+0.1)/(1-0.1)]=0.1003

x4=1/2*log[(1+0.04)/(1-0.04)]=0.04002

M=1/2*log[(1+0.07)/(1-0.07)]=0.0701

SD=1/sqrt(540-3)=>0.043

P(X 0.758761928

P(X 0.24210891

The probability is => P(X 0.516653

12. Applying Fisher transformation using the below expression:

=> 1/2*(log((1+r)/(1-r))

Standard error:

=>1/sqrt(n-3)

Lets find the values for 19% and 27%

x19=1/2*log[(1+0.19)/(1-0.19)]=0.1923

x27=1/2*log[(1+0.27)/(1-0.27)]=0.2769

M=1/2*log[(1+0.23)/(1-0.23)]=0.2342

SD=1/sqrt(602-3)=>0.0409

P(X>x27)=> 1-P(X 1-0.85176=>0.14824

P(X 0.152811

The probability is => P(X>x27) + P(X0.301

13. CI= mean ± tn-1 (sigma/sqrt(N))

Mean=3.9, sigma=0.8, N=208,CI=0.8

Since it’s a two-sided interval,

Therefore, t208-1 for (1-0.8)/2 will be=0.920441

=>3.9 + tn-1 (sigma/sqrt(N)) => 3.9 + 0.920441*(0.8 / sqrt(208))=>4.0=Upper bound

=>3.9 + tn-1 (sigma/sqrt(N)) => 3.9 - 0.920441*(0.8 / sqrt(208))=>3.9=lower bound

14. CI= mean ± tn-1 (sigma/sqrt(N))

Mean=16.6, sigma=11, N=7472,CI=0.98

Since it’s a two-sided interval, 0.1905

Therefore, t208-1 for (1-0.98)/2 will be= 0.992022

=>16.6 + tn-1 (sigma/sqrt(N)) => 16.6 + 0.992022*(16.6 / sqrt(7472))=>16.8=Upper bound

=>16.6 + tn-1 (sigma/sqrt(N)) => 16.6 - 0. 992022*(16.6 / sqrt(7472))=>16.4=lower bound

15. The figure in the top tight corner is the correct depiction of the problem.

Area under curve = 0.05, deg of freedom=26

Therefore using value of corresponding t=P-1 (0.05, 26) => -1.70562

16. sample = 383.6, 347.1, 371.9, 347.6, 325.8, 337

Mean=352.17

SD=21.68

Critical value = z0.05 =>-1.645

Lower bound = 352.17 - 1.645 * (21.68/sqrt(6))=>337.61

Upper bound= 352.17 + 1.645 * (21.68/sqrt(6))=>366.73

17. Mean=46.4 bushels

SD=2.45 bushels

Critical value = z0.10 =>-1.282

Lower bound = 46.4 - 1.282 * (2.45/sqrt(16))=> 45.62

Upper bound= 46.4 + 1.282 * (2.45/sqrt(16))=> 47.19

18. Mean=8

SD=1.9

Critical value = z0.005 =>-2.5758

Sample size=(2.5758 * 1.9/0.13)^2=>1417

19. Mean=12.6

SD=sqrt(3.61)=>1.9

Critical value = z0.025 =>-1.95996398

Sample size=(1.95996398* 1.9/0.19)^2=>384

20. Total sample of tenth graders = 2089, sample that reads above 8th grade level=1734

Therefore, proportion of tenth graders reading at or below the eighth grade level=(2089-1734)/ 2089=>0.170

Using the margin of errors formula.

Margin Of Error => z98 * sqrt((.17)*(1-.17)/2089)

Therefore, upper bound = 0.17 + 2.32634787sqrt((.17)(1-.17)/2089)=0.189

Lower bound=0.17 - 2.32634787sqrt((.17)(1-.17)/2089)=0.151

21. Total sample of tankers = 474, sample that had spills =156

Therefore, proportion of tankers that had spills =(474-156)/ 474=>0.671

Using the margin of errors formula.

Margin Of Error => z0.025 * sqrt((0671)*(1-0.671)/474)

Therefore, upper bound = 0.671 + 1.959963985sqrt((0.671)(1-0.671)/474=0.287

Lower bound=0.671 - 1.959963985sqrt((0.671)(1-0.671)/474=0.371