A cement truck delivers mixed cement to a large construction site. Let x represent the cycle time in minutes for the truck to leave the construction site, go back to the cement plant, fill up, and return to the construction site with another load of cement. From past experience, it is known that the mean cycle time is m = 45 minutes with s = 12 minutes. The x distribution is approximately normal.

(a) What is the probability that the cycle time will exceed 60 minutes, given that it has exceeded 50 minutes?

(b) What is the probability that the cycle time will exceed 55 minutes, given that it has exceeded 40 minutes?

#### Solution

Let x represent the cycle time in minutes for the truck to leave the construction site.

Given that $\mu = 45$ minutes and $\sigma = 12$ minutes.

a) The probability that the cycle time will exceed 60 minutes, given that it has exceeded 50 minutes is

` $$ \begin{aligned} P(X > 60 |X > 50) &= \frac{P(X > 60, X > 50)}{P(X > 50)}\\ &= \frac{P(X > 60) }{P(X > 50)}\\ &= \frac{P\bigg(\frac{X-\mu}{\sigma} > \frac{60-45}{12}\bigg)}{P\bigg(\frac{X-\mu}{\sigma}>\frac{50-45}{12}\bigg)}\\ &= \frac{P\big(Z > 1.25\big)}{P\big(Z > 0.416\big)}\\ &= \frac{1-P\big(Z < 1.25\big)}{1-P\big(Z < 0.416\big)}\\ &= \frac{1-0.8943502}{1-0.6616608}\\ &= 0.3122599. \end{aligned} $$ `

b) The probability that the cycle time will exceed 55 minutes, given that it has exceeded 40 minutes is

` $$ \begin{aligned} P(X > 55 |X > 40) &= \frac{P(X > 55, X > 40)}{P(X > 40)}\\ &= \frac{P(X > 55) }{P(X > 40)}\\ &= \frac{P\bigg(\frac{X-\mu}{\sigma} > \frac{55-45}{12}\bigg)}{P\bigg(\frac{X-\mu}{\sigma} > \frac{40-45}{12}\bigg)}\\ &= \frac{P\big(Z > 0.833\big)}{P\big(Z > -0.416\big)}\\ &= \frac{1-P\big(Z < 0.833\big)}{1-P\big(Z < -0.416\big)}\\ &= \frac{1-0.7975776}{1-0.3383392}\\ &= 0.3059307. \end{aligned} $$ `