91影视

Mean is$渭=0.37$

Standard deviation is$蟽=0.04$

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.

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

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.

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 .

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.