Question

I think I can! An important measure of the performance of a locomotive is its “adhesion,” which is the locomotive’s pulling force as a multiple of its weight. The adhesion of one $4400$-horsepower diesel locomotive varies in actual use according to a Normal distribution with mean $\mu =0.37$and standard deviation .$\sigma =0.04$For each part that follows, sketch and shade an appropriate Normal distribution. Then show your work.

(a) For a certain small train’s daily route, the locomotive needs to have an adhesion of at least $0.30$for the train to arrive at its destination on time. On what proportion of days will this happen? Show your method.

(b) An adhesion greater than $0.50$for the locomotive will result in a problem because the train will arrive too early at a switch point along the route. On what proportion of days will this happen? Show your method.

(c) Compare your answers to (a) and (b). Does it make sense to try to make one of these values larger than the other? Why or why not?

Short answer

a) We would expect trains to arrive about $96\%$of the time.

b) We would expect trains to arrive early about $0.06\%$of the time.

c) it makes sense to try to have the value found in part (a) larger.

Step-by-step solution

Part (a) Step-1 Given Information

Mean is$\mu =0.37$

Standard deviation is$\sigma =0.04$

Part (a) Step-2 Explanation

It is assumed that for the locomotive to arrive on time at its destination, it requires at least $0.30$adhesion.

Therefore, we want to find the place to the right of $0.3$in the $N(0.37,0.04)$distribution.

Find the value of $z-$corresponding to

$$\begin{gathered} z=\frac{0.3-0.37}{0.04} \\ =-1.75 \end{gathered}$$

In the following graph, the area above $0.3$is indicated. In the following graph, the area above $0.3$is indicated.

The value below $-1.75is1.0401$from the standard normal table,

Then, the area above $-1.75is1-0.0401=0.9599$.

Accordingly,$96\%$of the time, trains should arrive on time.

Part (b) Step-1: Given Information

An adhesion greater than $0.50$for the locomotive will result in a problem because the train will arrive too early at a switch point along the route

Part (b) Step-2: Explanation

Determine the z-value corresponding to $0.5$as follows:

$$\begin{gathered} z=\frac{0.5-0.37}{0.04} \\ =3.25 \end{gathered}$$

An area above $0.5$is equal to an area to the right of $0.25$. The following graph shows the area above $0.5$

The standard normal table shows that the proportion of adhesions over $0.50is1-0.9994=0.0006$

Hence trains to be expected to arrive early about $0.06\%$of the time.

Part (c) Step-1: Given Information

We have to compare (a) and (b) in order to know Does it make sense to try to make one of these values larger than the other .

Part (c) Step-2: Explanation

We want the train to arrive at its destination on time, but we don't want it to arrive too early at the turnaround point.