Question

Using information from the picture, what is the longest time required to pack and box one system with probability 0.95?

Packaging Stereos Consider the company that manufactures and ships small stereo systems that we discussed previously If the time required to pack the stereos can be described by a Normal distribution, with a mean of 9 minutes and standard deviation of 1.5 minutes and the times for the boxing stage can also be modeled as Normal, with a mean of 6 minutes and standard deviation of I minute, what is the probability that packing an order of two systems takes over 20 minutes? What percentage of the stereo systems takes longer to pack than to box?

Answer 1

Let X and Y be the random variables denoting the time to pack and box the stereo respectively.

Then X ~ N(9, 1.5²) and Y ~ N(6, 1²)

Total time to pack 2 stereos = 2X

Probability that the packing an order of two systems takes longer than 20 minutes = P(2X > 20)

= P(X > 10) = P[Z > (10 - 9) / 1.5] = P[Z > 0.6667]

= 0.2525

Percentage of stereo systems takes longer to pack than to box = P(X > Y)

= P(X - Y > 0)

Now,

Now, E(X-Y) = E(X) - E(Y) = 9 - 6 = 3 and Var(X-Y) = Var(X) + Var(Y) = 1.5² + 1²  = 3.25

Standard deviation of X - Y = $\sqrt{3.25}$ = 1.803

Thus, X - Y ~ N(3, 1.803²)

Percentage of stereo systems takes longer to pack than to box = P(X > Y)

= P(X - Y > 0) = P[Z > (0 - 3) /1.803] = P[Z > -1.664]

= 0.9519

= 95.19%

Time to pack and box one system is X + Y

Now, E(X+Y) = E(X) + E(Y) = 9 + 6 = 15 and Var(X+Y) = Var(X) + Var(Y) = 1.5² + 1²  = 3.25

Standard deviation of X + Y = $\sqrt{3.25}$ = 1.803

Thus, X + Y ~ N(15, 1.803²)

Z value for probability 0.95 is 1.645

The longest time required to pack and box one system with probability 0.95 = Mean - z * SD

= 15 - 1.645 * 1.803 = 12.034 minutes