Chapter 2.2, Problem 57E

(a)

To determine

To find: the value of the mean adhesion of the locomotive

Answer to Problem 57E

The mean adhesion should be 0.382

Explanation of Solution

Given:

The standard deviation remains at $\sigma =0.04$ ,

Calculation:

Solve for the appropriate mean of the distribution so that the adhesion is less than 0.30 on less than $2\%$ of days.

Find the z -value for which $2\%$ of observations is lower

Using Standard normal table, the z -value corresponding to 0.02, $z=-2.05$ .

Since this z -value should correspond to an adhesion of 0.30,

  $\begin{array}{l}-2.05=\frac{0.30-\mu }{0.04}\\ \mu =0.382\end{array}$

Hence, the mean adhesion should be 0.382

Conclusion:

Hence, the mean adhesion should be 0.382

(b)

To determine

To find: the value of the standard deviation of the adhesion

Answer to Problem 57E

The standard deviation of the adhesion values should be 0.034.

Explanation of Solution

Given:

The mean adhesion stays at $\mu =0.37$

Calculation:

If the mean adhesion stays at 0.37, find the standard deviation that is to be decreased to ensure that the train will arrive late less than $2\%$ of the time.

To find the standard deviation,

  $\begin{array}{l}-2.05=\frac{0.30-0.37}{\sigma }\\ \sigma =0.034\end{array}$

The standard deviation of the adhesion values should be 0.034.

Conclusion:

Therefore, the standard deviation of the adhesion values should be 0.034.

(c)

To determine

To find: whether option a or b is preferable.

Answer to Problem 57E

Part (b) is preferred.

Explanation of Solution

Given:

The mean adhesion stays at $\mu =0.37$

Calculation:

Compare the two options in (a) and (b) to find which option is preferable. Our objective is to decrease the variability in adhesion from trip to trip.

To compare the options, find the area under the $N\left(\mu ,\sigma \right)$ to the right of 0.50. Taking the standard deviations as 0.04, calculate the z value as follows:

  $\begin{array}{l}z=\frac{0.50-0.382}{0.04}\\ =2.95\end{array}$

Using standard normal table, find that the area to the right of 0.5 as follows:

  $1-0.9984=0.0016$

That is, using $\sigma =0.04$ the proportion of objects that are to right of 0.5 are 1.6%

Taking the standard deviations as 0.034, calculate the z value as follows:

  $\begin{array}{l}z=\frac{0.50-0.37}{0.034}\\ =3.82\end{array}$

Using standard normal table, find that the area to the right of 0.5 as follows:

  $1-0.9999=0.0001$

That is, using $\sigma =0.034$ , the proportion of objects that are to right of 0.5 are 0.01%

Since very less percentage of proportions isto the right of 0.5 when $\sigma =0.034$ used, part (b) is preferred.

Conclusion:

Therefore, part (b) is preferred.