Question

Errors in measuring truck weights. To help highway planners anticipate the need for road repairs and design futureconstruction projects, data are collected on the estimatedvolume and weight of truck traffic on specific roadways (Transportation Planning Handbook, 2016) using specialized “weigh-in-motion” equipment. In an experiment involving repeated weighing of a 27,907-pound truck, it wasfound that the weights recorded by the weigh-in-motionequipment were approximately normally distributed witha mean of 27,315 and a standard deviation of 628 pounds

(Minnesota Department of Transportation). It follows thatthe difference between the actual weight and recordedweight, the error of measurement, is normally distributedwith a mean of 592 pounds and a standard deviation of 628pounds.

a. What is the probability that the weigh-in-motion equipment understates the actual weight of the truck?

b. If a 27,907-pound truck was driven over the weigh-in-motion equipment 100 times, approximately howmany times would the equipment overstate the truck’sweight?

c. What is the probability that the error in the weightrecorded by the weigh-in-motion equipment for a27,907-pound truck exceeds 400 pounds?

d. It is possible to adjust (or calibrate) the weigh-in-motion equipment to control the mean error of measurement. At what level should the mean error beset so the equipment will understate the weight of a27,907-pound truck 50% of the time? Only 40% of thetime

Short answer

a. The probability that the weigh-in motion equipment understates the actual weight of the truck is 0.8271

b. The equipment understates the actual weight of the truck0.1729 times

c. The probability that the error in the weight recorded by the weigh-in motion equipment for a 27,907 pound truck exceeds 400 poundsis 0.677

The mean error should be set so the equipment will understate the weight of a 27,907-pound truck 50% of the timeat 0levelsand 40% of the time at -159.0742 level

Step-by-step solution

Given information

X be the weight measurement by the equipment,

X follows a normal distribution with a mean 27315 and a standard deviation 628

Calculating the probability of the weigh-in-motion equipment understating the actual weight of the truck.

a.

given $X~N(mean=27315,sd=628)$

to calculate the probability of the weight in motion equipment understate the actual weight of the truck that is to find $\mathrm{P}\left(understate\right)$

Hence,

$$\begin{gathered} \mathrm{P}\left(understate\right)=P\left(X\le 27907\right) \\ =P\left(Z\le \frac{X-\mu }{\sigma }\right) \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(understate\right)=\left(Z\le \frac{27907-27315}{628}\right) \\ =\left(Z\le \frac{592}{628}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(understate\right)=P\left(Z\le 0.942\right) \\ =0.8271 \end{gathered}$$

Hence,the probability that the weigh in motion equipment understates the actual weight of the truck is 0.8271

Calculating the number of times the equipment understates the actual weight of the truck.

b.

To calculatethe equipment understates the actual weight of the truck that is to calculate

$$\begin{gathered} \mathrm{P}\left(overstate\right)=\mathrm{P}\left(X>27907\right) \\ =1-\mathrm{P}\left(X\le 27907\right) \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(overstate\right)=1-0.8271 \\ =0.1729 \end{gathered}$$

Hence, the equipment understates the actual weight of the truck 0.1729 times

Calculating the probability that the error in the weight recorded by the weigh-in motion for 27,907 pound truck exceeds 400 pounds

c.

Given that E follows a normal distribution with $mean=592$and standard deviation is 628

$$\begin{gathered} P\left(errorexcuted400\right)=\mathrm{P}\left(E>400\right)+\mathrm{P}\left(E<-400\right) \\ =\mathrm{P}\left(Z>\frac{X-\mu }{\sigma }\right) \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(errorexceed400\right)=P\left(Z>\frac{400-592}{628}\right)+P\left(Z<\frac{-400-592}{628}\right) \\ =0.67722 \end{gathered}$$

Therefore, the probability that the error in the weight recorded by the weigh in motion for 27907 pound truck exceeds 400 pounds is 0.677.

Calculating the mean error rate.

d.

50% of time: Suppose new error mean to be set is m

$$\begin{gathered} \mathrm{P}\left(understate\right)=P\left(E>0\right) \\ =P\left(Z>\frac{x-\mu }{\sigma }\right) \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(understate\right)=\mathrm{P}\left(Z>\frac{0-m}{6288}\right) \\ =0 \end{gathered}$$

Hence$m=0$

Hence The mean error be set so the equipment will understate the weight of a 27,907 pound truck 50% of time at -0 level.

40% of time: Suppose new error mean to be set is m

$$\begin{gathered} \mathrm{P}\left(understate\right)=P\left(E>0\right) \\ =\mathrm{P}\left(Z>\frac{x-\mu }{\sigma }\right) \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(understate\right)=\mathrm{P}\left(Z>\frac{0-m}{6288}\right) \\ =0.40 \end{gathered}$$

$$\begin{gathered} P\left(Z>0.2533\right)=0.4 \\ =\frac{0-\mathrm{m}}{628} \\ =0.2533 \end{gathered}$$

Hence$m=-159.0724$

The mean error be set so the equipment will understate the weight of a 27,907 pound truck 40% of time at -159.0724 level.