91影视

X be the weight measurement by the equipment,

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

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,

$$\begin{gathered} \mathrm{P}\left(understate\right)=P\left(X鈮�27907\right) \\ =P\left(Z鈮�\frac{X-渭}{蟽}\right) \end{gathered}$$

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

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

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

b.

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

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

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

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

c.

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

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

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

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.

d.

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

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

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

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

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

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

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

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.