91影视

The standard deviation remains at $蟽=0.04$

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

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" $$\begin{gathered} -2.05=\frac{0.30-渭}{0.04} \\ 渭=0.382 \end{gathered}$$

Hence, the mean adhesion should be $0.382$

Mean adhesion stays at$渭=0.37$

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

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,

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

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

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.

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 $N(渭,蟽)$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 $蟽=0.034$, 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.