Statistics Assignment: Probability, Bayes, and Distributions

This statistics assignment addresses several key concepts in probability and statistics. It begins with probability calculations involving damaged fruits from different countries and then explores conditional probability using Bayes' theorem in a scenario involving defective springs from different machines. The assignment covers binomial distributions, calculating probabilities, and expected values. It also delves into geometric distributions, calculating probabilities of success, and the concept of expected value in this context. Further, the assignment explores Poisson distributions and gamma distributions, deriving equations and performing calculations. The assignment concludes with an analysis of normal distributions, including transformations and their impact on the distribution parameters. Throughout the assignment, various formulas and theorems are applied to solve the problems, providing a comprehensive overview of statistical principles.

SUBJECT: STATISTICS
ORDER ID: 62758
Q1.
Note that Q1 and Q2 does not apply Bayes Theorem but this concept is applicable in SQ3
(a)
Probability of getting a box from Thailand = 6000/10000=3/5
Probability of getting a box from Malaysia=4000/10000=2/5
Probability of getting a box from Thailand that is damaged=200/6000=1/30
Probability of getting a box from Malaysia that is damaged=365/4000=73/800
Therefore the probability of selecting a box at random and get a damaged fruit = (3/5*1/30) +
(2/5*73/800)
=113/2000
(b)Probability of selecting a box at random that comes from Malaysia=2/5
Probability of selecting a box at random with an overripe fruit that comes from
Malaysia=295/4000=59/800
Thus the probability=2/5*59/800=59/2000
Q2.
25% emits excessive pollutants,75% does not emit excessive pollutants
0.95 is the probability that a car emitting excessive pollutants fail
0.1 is the probability that a car not emitting excessive pollutants fail
Probability of a car emitting excessive pollutants and fails the test is
=0.25*0.95=0.2375
Q3.
Probability of selecting any machine at random=1/3

Machine 1 defective springs=0.02, produces 0.35 of the total springs
Machine 2 defective springs=0.01, produces 0.25 of the total springs
Machine 3 defective springs=0.03produces 0.4 of the total the springs
(a)
Probability of selecting a defective spring=(1/3*0.02)+(1/3*0.01)+(1/3*0.03)=1/50
(b)
Probability of(3/defective)=P(3)*P(D/3) divided by P(3)*P(D/3)+P(1)*P(1/D)+P(2)*P(2/)
Where d= defective
(0.03*0.04)/P(0.03*0.04)+(0.02*0.35)+P(0.01*0.25)
=1.2*10^-3/0.0107=0.11215
This is where Bayes theorem applies but question 1 and 2 are not using Bayes Concept
Q4.
q^(n-x)*p^x
p=0.15,q=1-p,1-0.3
Q=0.7
(a)
P(x=5)
0.7^(15-5)*0.3^5
15C5*0.7^10*0.3^5(15C5 all I mean is 15 combination )
=0.2061
(b)
P(x<_5)

(15C0multiplied by 0.7^15multiplied by 0.3^0)+ (15C1multiplied by 0.7^14multiplied by 0.3^1) + (15C2o
the product of 0.7^13o the product of 0.3^2) + (15C3multiplied by 0.7^12multiplied by 0.3^3) +
(15C4multiplied by 0.7^11multiplied by 0.3^4) +0.2061
=0.00475+0.0305+0.0916+0.1700+0.2186+0.2061
=0.72155
(c)
E(X)=np
=E(X^2) =np^2
E(X) =0.3*15=4.5
E(X^2) =0.3^2*15=1.35
Q5.
P(A=a) = (0.2)*(0.8^(x-1))
Where a=1, 2, 3
a=0.2*0.8^ (1-1) +0.2*0.8^ (2-1) +0.2*0.8^ (3-1)
=0.2+0.16+0.128
=0.488
Y=X-1
X=0.488
x=0.2*0.8^1.1
=0.2
=0.488-0.2
=0.288
Q6.
(a)
1-p=q

Thus Z* (g-1)multiplied by g=1, 2,3
Where Q=Z and g=n
E (L) =1/p= 1/0.3*2=6.6667
Var (L) =q/0.09*2
=0.7/0.09*2
=15.556
E (N)=1/p=1/0.3=3.33
Var (N)=q/p^2=0.7/0.09=7.778
(b)
=P (N>j)
Let k=j=0, 1,2,3,4,
Therefore P multiplied by (N greater than k+j|N>k) =P (N>j)
Then P multiplied to (N>j) since j=0, 1, 2, 3, 4
P (N>j) =P (N>k+j/N>k)
P (N>k+j-k)
P (N>j) =P (N>j)
Q7
=lambda1Ce*lambda/K1 k=0, 1, 2(please write letter y that is upside down for lambda, C is for
combination like in Q4)
Thus –lambdaCe+-lambdaCe*lambda1+-lambda*lambda2/2….. (Equationi)

For (P(X2=k) =N2k/k*e*lambda2 for k=0, 1, 2
E*-lambda*lambda2/k for k=0, 1, 2
-lambdaCe+-lambda2Clambda+-lambda2*lambda2/2… (Equation i)

Since lambda is a constant of proportionality rate of parameter which is equation i then we substitute it
with equation ii …e( to the power of -1)+e(raised to-1*1)+e(raised to-1*0.5)….(eqn i)

e^-1 +e^-1*1+e^-1*1/2…eqn ii
from eqn i we get 0.368+0.368+0.184=0.92
from eqn ii we get 0.368+0.368+0.184=0.92
=0.92*2
=1.84
Q8.
By the help integration and given that j is greater than zero and i is less zero then the parameters by
which the variable lies are
Change in y/ change in x=AdB/dx+BdA/dx his where A and B represent U and V as the variables
Change in y divided by change=
1multplied by e raised to power -1*(k-1)!divided by k
Integral of d UV) dx/dx=integral of Udv/dx*dx integral of Vdu/dx*dx
So in regard to the query U=1 and V= (k-1)! 0^k^-1*e-1/0
Integral of infinity to infinity (-1*e-1/1*(k-1)! /k
Given that X1~lambda and x2~lambda then from the above it is evident that X1+x2~gamma (2,1lambda)
Q10.
X~N (u, d^2)
Taking A and b for and b for and substitute it for the equation below we get;
X~N (A, d^2) = X~NA, Nd^2 on the other side taking N (Au+b, A^2d^2) hence
Ax+b where b=d^2 and X=N (Au+b, A^2*d^2
Thus AX+b ~N(Au+b,A^2*d^2)

Please note b stands for beta and d for sigma
Q9
X1,…Xn~lambda;
Given the fact that the variables are mutually independent and random then to find the then
possible smallest limit of X we say;
Min (X1,…Xn)exponential ( the integral of postive infinity to n taken to the product of
lambda and b)
Proofing the equatiion above
Fk(K)=A(X1<_x)
=1-raised to power negative lambda multplied by variable K and contant b squared.
References
a) Janet, S.Abbot (1987) Mathematics today.Orlando, Harcourt Brace Jovanovich Publishers.
b) Duncan,C(1979) Psychophysics and psychology hearing Vol. 93.Hillsdale,NJ Lawrence Erlbaum
Associates.