Question

A company manufactures small stereo systems. At the end ofthe production line, the

stereos are packaged and prepared for shipping. Stage 1of this process is called
"packing" (collecting al the components, putting everything ni plastic bags, etc.). Stage
2 of this process si called "boxing" (arranging everything in the box, sealing the box,
labeling It, e t c . .t h e company says that times required for packing are approximately
normally distributed with a mean of 9 minutes and a SD of 1.5 minutes. The times for
boxing are also approximately normal, with a mean of 6 minutes and a SD of 1 minute.



What is the probability that “packing” two consecutive systems takes over 19 mins?

Answer 1

0.32, or 32%.

Step-by-step explanation:

We can model the time it takes to pack a single stereo system using a normal distribution with a mean of 9 minutes and a standard deviation of 1.5 minutes. Let X be the time it takes to pack a single stereo system, then X ~ N(9, 1.5^2).

To find the probability that packing two consecutive systems takes over 19 minutes, we need to find the distribution of the sum of two independent normal random variables, which is also a normal distribution. Let Y be the total time it takes to pack two consecutive stereo systems, then Y = X1 + X2, where X1 and X2 are independent and identically distributed as X.

The mean of Y is:

E(Y) = E(X1 + X2) = E(X1) + E(X2) = 9 + 9 = 18

The variance of Y is:

Var(Y) = Var(X1 + X2) = Var(X1) + Var(X2) = 1.5^2 + 1.5^2 = 4.5

The standard deviation of Y is:

SD(Y) = sqrt(Var(Y)) = sqrt(4.5) ≈ 2.12

So, Y ~ N(18, 2.12^2).

Now we can find the probability that packing two consecutive systems takes over 19 mins by standardizing the distribution of Y and calculating the corresponding z-score:

P(Y > 19) = P(Z > (19 - 18) / 2.12) ≈ P(Z > 0.47)

Using a standard normal distribution table or calculator, we can find that P(Z > 0.47) ≈ 0.32.

Therefore, the probability that packing two consecutive systems takes over 19 minutes is approximately 0.32, or 32%.