Question

A company manufactures Printed Circuit Boards (PCBs) expects to have 6 defective units each day. Let Y be a random variable that counts the number of defective units produced each day.

A. Which discrete random variable distribution would best model this scenario?

B. What is the probability that the number of defective units observed in a dayexceeds the mean number by more than one standard deviation?

C. What is the probability that, on two randomly selected days, no defective units are observed?

Answer 1

Solution:-

Given

A company manufactures Printed Circuit Boards (PCBs) expects to have 6 defective units each day. Let Y be a random variable that counts the number of defective units produced each day.

a) Which discrete random variable would best model this scenario?

We know that Poisson distribution is the discrete probability distribution of the number of events occuring in a given time period, given the average number of times the event occurs over that time period.

So, Poisson distribution would best model this scenario.

b) What is the probability that the number of defective units observed in a day exceeds the mean number by more than one standard deviation?

So,

Now, we know that, variance of Poisson distribution is the mean itself.

Var (y) = 6

Standard deviation (Y) = 2.449

One standard deviation of mean is (6 + 2.449) = 8.449

Required probability: P(Y > 8.449)

Again,

  

= 1 - 0.847

= 0.153

c) What is the probability that, on two randomly selected days,no defective units are observed?

Now, the events of observing no defecting units on two days are independent.

Required probability :

= 0.000004

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