Question about probabilities and statistics....I need help!!!?

The amount of time, in hours, that a computer functions before breaking down is continuous random variable with exponential density function:





f(x)= 0.01exp(-x/100) , x %26gt;=0


f(x)=0 , x%26lt;0





a) What is the probability that a computer will function more than 100 hours before breaking down?


b) If there are 4 such computers. What’s the probability that at most 3 computers will function more than 100 hours before breaking down?





(second problem)


A bakery uses 3000 raisins to make 100 loaves of bread. Let x be the number of raisins in each loaf. X follows binomial distribution with n=3000, p=1/100. Using central limit theorem (normal approximation):


a) Find the probability p(28%26lt;x), p(x%26gt;32)


b) Find the probability that average number of raisins in 36 randomly chosen loaves exceeds 32.


c) Compare results in (a) and (b)


d) Why is it possible to apply central limit theorem to this problem?


@ Find a and b such that p(a%26lt;x%26lt;b)= .99

Question about probabilities and statistics....I need help!!!?
Q1


X has exp distribution and so the distribution function is


P(X %26lt;= x) = F(x) = 1 - exp(-x/100)





a) We find that P(X %26lt;= 100) = 1 - exp(-1)


and P(X %26gt; 100) = exp(-1)





b) Y = the number of computers (out of 4) that work 100 h or more


We have Y ~ BIN(n = 4, p = exp(-1))


and


P( Y %26lt;= 3)


= 1 - P(Y = 4)


= 1 - p^4


= 1 - exp(-4)





Q2


X ~ BIN(n = 3000, p = 1/100)


The normal approximation gives


X ~~ N(np, var = np(1-p)) = N(30, var = 29,7)


(a1) P(28 %26lt; X)


= (cont. correction) P(28,5 %26lt; X)


= (standardize) P( (28,5 - 30)/sqrt(29,7) %26lt; (X - 30)/sqrt(29,7))


= (CLT) P( -0, 275 %26lt; Z)


= 0,608


(a2) P(X %26gt; 32)


= P(X %26gt; 32,5)


=..


= P(Z %26gt; 0,458)


= 0,323


(b) average of m = 36 loaves is approx N(np, var = np(1-p)/m)


P(average %26gt; 32)


~ P( Z %26gt; (32 - 30)/sqrt(29,7/36))


= P(Z %26gt; 2,20)


= 0,014


(c) In b we find a smaller prob because the variance of the average = the variance divided by 36


(d) because n = 3000 is large





(in my opinion, a Poisson approximation would also work)





(e) There are many solutions. I look here for a symmetric solution of the form a = 30 - c and b = 30 + c


P(30 - c %26lt; X %26lt; 30 + c) = 0,99


or (after standardizing and approximating)


P(-c/sqrt(29,7 %26lt; Z %26lt; +c/sqrt(29,7)) = 0,99


and then


P(Z %26lt; c/sqrt(29,7))= 0,995


We find c/sqrt(29,7) = 2,58


and c = 2,58*sqrt(29,7) = 14,06





So, 1 answer is that


P(30 - 14,06 %26lt; X %26lt; 30 + 14,06) = 0,99