1. (40 pts) A factory produces Xn units of output on day n where the Xn are IID

Question

1. (40 pts) A factory produces Xn units of output on day n where the Xn are IID (i.e., independent and identically distributed RVs) with unknown shape but with mean () = 5 and variance (2 ) = 25

a. (10 pts) What is the approximate probability that the cumulative sum of output produced in 100 days is more than 600? (Hint: You may want to use the Central Limit Theorem. Also assume that the Xn are independent random variables)

b. (10 pts) What is the approximate probability that the cumulative sum of output produced in 100 days is between 400 and 600 units?

c. (10 pts) Find the largest value of N such that: P((x1 +…+ xN) (300 + 5N) = 0.05

d. (10 pts) Let N675 be the 1st day on which the total number of gadgets produced exceeds 675.

i. Calculate an approximation to the probability that N675 150.

ii. Calculate an approximation to the probability that N675 135.

Explanation / Answer

a) P(X1 +X2 ... X100 > 600)

S = X1 + ..Xn

E(S) = n*

sd(S) = sqrt(n) *

here n = 100 , = 5 , = 5

E(S) = 500 , sd(S) = 50

Z= (S - 500)/50

P(S > 600)

P(Z>(600 -500)/50)

=P(Z > 2) = 0.0228

b) P(400 <S< 600)

=P ( 2<Z<2 )=0.9544

c) P(s > (300 + 5n)) = 0.05

P(Z >(300 +5n - 5*n)/(5*sqrt(n)) = 0.05

P(Z> (60/sqrt(n)) = 0.05

P(Z> z*) = 0.05

z* = 1.645

hence (60/sqrt(n) = 1.645

n < 1330.36

n = 1330

d) P(S > 675)

=P(Z >(675 - 5*n)/(5*sqrt(n))

P(N <= 150) = P(Sn > 675 in 150 days)

= P(Z > ((675 - 5*150)/(5*sqrt(150))

= P(Z> -1.2247)

= 0.8897

P(N>=135) = 1 - P(Sn > 675 in 134 days)

= 1 - P(Z >((675 - 5*134)/(5*sqrt(134))

= 1 - P(Z> 0.08638684255)

= 0.5344

Please don't forget to rate positively if you found this response helpful.
Feel free to comment on the answer if some part is not clear or you would like to be elaborated upon.
Thanks and have a good day!

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.