Ma260 Statistical Analysis Assignment

MA260 Statistical Analysis I

Lesson 6 Assignment 06

August 17 2012

1. Mean = E(x) = Sum (xP(x)) = 2(0.5) +8(0.3) +10(0.2) = 5.4

Ans: Mean = 5.4

Variance = V(x) = E(x2) - E(x)2

E(x2) = Sum (x2P(x)) = 22(0.5) +82(0.3) +102(0.2) = 2+19.2+20 = 41.2

V(x) = 41.2 – (5.4)2 = 12.04

Ans: Variance is 12.04

2. Total number of cases are: 8C3 = 56 , Favorable cases are 6C3 =20

P (all members are tenured) = 20/56 = 5/14

b. P (at least one member not tenured) = 1-P (all members tenured) = 1-5/14 = 9/14

3. Let X = number of orders unfilled , X has a Poisson distribution with mean λ=2

P(X=k) =2ke-2k!

P(X fewer than 5) = P(X<5) = P(X=0) +P(X=1) +P(X=2) +P(X=3) +P(X=4)

= 20e-20!+21e-21!+22e-22!+23e-23!+24e-24!=0.9473

So yes, it is reasonable to conclude that 95% of the working days there is fewer than 5 orders unfilled.

4. Let X= number of Hondas in the sample of three

Range of X = {0, 1, 2, 3}

P(X=0) = 5C3/9C3 = 10/84= 5/42

P(X=1) = (4C1) (5C2)/9C3 = 40/84 = 10/21

P(X=2) = (4C2) (5C1)/9C3 = 30/84 = 5/14

P(X =3) = 4C3/9C3 = 4/84 = 1/21

Answer:

Probability distribution:

X P(x)

0 5/42

1 10/21

2 5/14

3 1/21

b. P (At least one Honda is included) = P (X≥1) = 1-P(X=0) =1-p (0) = 1-5/42 = 37/42

Ans: 37/42

5. If there is no truth in this theory (probability are the same)

Then P (up in the year) =P (down in the year) = 0.5

Let X = number of times that the theory is right in 34 years

X has a binomial distribution with n = 34 and p = 0.5

P(X=29) = (34C29) (0.5)29(0.5)5 = 278256(0.5)34= 0.000016197

Ans: 1.6197x10-5