Question

You are a cook, you have to check the baked meat every 3 hours. In each 3 hour there is 1/2 probability it’s overcooked and independently cooked with probabllity 1/4.

a. If it is cooked or overcooked in a 3 hour span, the food needs to be pulled out. What is the expected number of times you have to pull out the meats in a week?

b. Approximate the probability you pulled out more than 50 meats in a given week? Express answer in terms of cumulative distribution function of a standard Normal random variable. P(X<= x), where X~N(0.1)

Answer 1

Probability that food is overcooked in 3 hours span = 1/2 = 0.5

Probability that food is cooked in hours span = 1/4 = 0.25

Hence probability that food is either overcooked or cooked = 0.5 + 0.25 = 0.75

Hence probability of pulling out meat in 3 hours span = 0.75

i.e. p = 0.75

a) We have to find expected number times to pull out the meat in a week.

1 week = 7 days = 168 hours

There a

re 56 periods of 3 hours in 1 week.

Thus n = 56

Distribution is Binomial with p = 0.75 and n = 56

Expected value:

Hence expected number of times you have to pull out the meats in a week is 42.

b)

Since n is larger and Normal approximation assumption np > 5 and np(1 - p) > 5 are met, we can approximate Binomial distribution as Normal distribution with

And

We have to find probability that more than 50 times meat is pulled out.

Let us find first.

Convert 'x' into z score.

Here here z is standard Normal random variable.

Cumulative distribution function is defined as F(x)