Question

Suppose that the duration of a particular type of criminal trial is known to have a mean of $21$days and a standard deviation of seven days. We randomly sample nine trials.

a. In words,$\Sigma X=\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

b.$\Sigma X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

c. Find the probability that the total length of the nine trials is at least $225$days.

d. Ninety percent of the total of nine of these types of trials will last at least how long?

Short answer

a. the total length of time for nine criminal trials

b.$N(189,21)$

c.$0.0432$

d. $162.09;$ninety percent of the total nine trials of this type will last $162$ days or more.

Step-by-step solution

Given information

Suppose that the duration of a particular type of criminal trial is known to have a mean of $21$days and a standard deviation of seven days. We randomly sample nine trials.

Explanation (part a)

definition for $\sum _{}X$: the total length of time for nine criminal trials

Explanation (part b)

The sum of random variables in normal distribution is given by, $\sum _{}X=N(n\mu ,\sqrt{n}\sigma )$

Plugging all the values in the above equation, we get,

$$\begin{gathered} \sum _{}X=N(9\times 21,\sqrt{9}\times 7) \\ \sum _{}X=N(189,21) \end{gathered}$$

Explanation (part c)

the probability that the total length of the nine trials is at least $225$days.

role="math" localid="1651602155511" $$\begin{gathered} P(\sum _{}X\le 225)=normalcdf(lower,upper,n\mu ,\sqrt{n}\sigma ) \\ P(\sum _{}X\le 225)=normalcdf(225,1E99,189,21)=0.0432 \end{gathered}$$

Explanation (part d)

Let$k=the90thprecentile.$

Find $k$, where $P(\sum _{}x<k)=0.90.$

$\text{invNorm}\left(0.90,189,21\right)=162.09$

ninety percent of the total nine trials of this type will last$162$days or more.