Question

I think I can! Refer to Exercise 55. The locomotive’s manufacturer is considering two changes that could reduce the percent of times that the train arrives late. One option is to increase the mean adhesion of the locomotive. The other possibility is to decrease the variability in adhesion from trip to trip, that is, to reduce the standard deviation.

(a) If the standard deviation remains at $\sigma =0.04$, to what value must the manufacturer change the mean adhesion of the locomotive to reduce its proportion of late arrivals to less than $2\%$of days? Show your work.

(b) If the mean adhesion stays at$\mu =0.37$, how much must the standard deviation be decreased to ensure that the train will arrive late less than $2\%$of the time? Show your worK

(c) Which of the two options (a) and (b) do you think is preferable?. justify your answer. (Be sure to consider the effect of these changes on the percent of days that the train arrives early to the switch point)

Short answer

a) The mean adhesion should be$0.382$

b) The standard deviation of the adhesion values should be $0.034$

c) We prefer part (b)

Step-by-step solution

Part (a) Step-1 Given Information 

The standard deviation remains at $\sigma =0.04$

The mean adhesion of the locomotive reduces its proportion of late arrivals to less than $2\%$of days.

Part (a) Step-2 Explanation

Therefore, the adhesion should be below $0.30$on less than $2\%$of days by finding the appropriate mean.

Firstly, we determine the $z-$value below which $2\%$of observations fall.

The $z-$value for $0.02,z=-0.05$, taken from the standard normal table.

This $z-$value corresponds to$0.30$adhesion, so we solve the following equation.

localid="1649932870412" $$\begin{gathered} -2.05=\frac{0.30-\mu }{0.04} \\ \mu =0.382 \end{gathered}$$

Hence, the mean adhesion should be $0.382$

Part (b) Step-1: Given Information

Mean adhesion stays at$\mu =0.37$

The standard deviation be decreased to ensure that the train will arrive late less than $2\%$of the time.

Part (b) Step-2: Explanation

In order for the train to arrive less than $2\%$late, the standard deviation must be decreased if the mean adhesion is$0.37$.

Solving the following equation will give us the standard deviation,

$$\begin{gathered} -2.05=\frac{.30-.37}{\sigma } \\ \sigma =0.034 \end{gathered}$$

The standard deviation of the adhesion values should be $0.034$.

Part (c) Step-1: Given Information

a) The mean adhesion should be$0.382$

b) The standard deviation of the adhesion values should be $0.034$we have to find that Which of the two options (a) and (b) do you think is preferable.

Part (c) Step-2: Explanation

The objective of this study is to decrease variance in adhesion from one trip to the next, so we will compare the two options shown in (a) and (b).

We calculate the $z$value by taking the standard deviations as $0.04$, and finding the area under the $N(\mu ,\sigma )$to the right of $0.50$:

localid="1649997605393" $$\begin{gathered} z=\frac{0.50-0.382}{0.04} \\ =2.95 \end{gathered}$$

According to the standard normal table, the area to the right of $0.5$appears like this:

$1-0.9984=0.0016$

Accordingly, $1.6\%$of the objects are to the left of $0.5$when $s=0.04$.

Calculating the $z$value with the standard deviations of $0.034$, we arrive at the following:

localid="1649997614065" $$\begin{gathered} z=\frac{0.50-0.37}{0.034} \\ =3.82 \end{gathered}$$

According to the standard normal table, the area to the right of$0.5$appears like this:

$1-0.9999=0.0001$

So, using $\sigma =0.034$, the objects proportion that are to right of $0.5are0.01\%$.

With $s=0.034$, very few percentages are to the right of $0.5$.

Hence, part (b) should be preferred.