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Question

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 μ=45\mu=45 minutes with σ=12\sigma=12 minutes. The x distribution is approximately normal. What is the probability that the cycle time will exceed 60 minutes, given that it has exceeded 50 minutes?

Solution

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Given:

μ=45\mu=45

σ=12\sigma=12

The z-score is the value decreased by the mean, divided by the standard deviation:

z=zμσ=604512=1.25z=\dfrac{z-\mu}{\sigma}=\dfrac{60-45}{12}=1.25

z=zμσ=5045120.42z=\dfrac{z-\mu}{\sigma}=\dfrac{50-45}{12}\approx 0.42

Rewrite the original interval as an interval of z-scores and then determine the corresponding probability using table 5 in the appendix.

P(x>60)=P(z>1.25)=1P(z<1.25)=10.8944=0.1056P(x>60)=P(z>1.25)=1-P(z<1.25)=1-0.8944=0.1056

P(x>50)=P(z>0.42)=1P(z<0.42)=10.6628=0.3372P(x>50)=P(z>0.42)=1-P(z<0.42)=1-0.6628=0.3372

Definition Conditional probability:

P(BA)=P(A and B)P(A)P(B|A)=\dfrac{P(A\text{ and }B)}{P(A)}

Replace AA by x>60x>60 and replace BB by x>50x>50

P(x>60x>50)=P(x>60 and x>50)P(x>50)=P(x>60)P(x>50)=0.10560.33720.3132P(x>60|x>50)=\dfrac{P(x>60\text{ and }x>50)}{P(x>50)}=\dfrac{P(x>60)}{P(x>50)}=\dfrac{0.1056}{0.3372}\approx 0.3132

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