Chapter 2.2, Problem 68E

(a)

To determine

60^th percentile of the first − class letter weights.

Answer to Problem 68E

60^th percentile is 0.73 ounces.

Explanation of Solution

Given information:

Mean, $\mu =0.69$

Standard deviation, $\sigma =0.16$

Calculations:

The data value represented by the x^th percentile has x% of the data values below it.

That implies

The data value represented by the 60^th percentile has 60% of the data values below it.

Now,

Find the z − score corresponding the probability of 60% (or 0.60) in the normal probability table of the appendix.

Note that

The closest probability would be 0.5987 that lies in the row 0.2 and in the column .05 of the normal probability table.

Then

The corresponding z − score,

  $z=0.25$

In order to find x^thpercentile, calculate the z − score

  $z=\frac{x-\mu }{\sigma }=\frac{x-0.69}{0.16}$

Equate the above two expressions:

  $\frac{x-0.69}{0.16}=0.25$

Multiply both sides by 0.16:

  $x-0.69=0.04$

Add 0.69 to both sides:

  $x-0.69+0.69=0.04+0.69$

That becomes

  $x=0.73$

Thus,

60^thpercentile will be 0.73 ounces.

(b)

To determine

Proportion of first − class letters at this post office that require extra postage.

Answer to Problem 68E

Around 0.0262 of the first − class letters at this post office require extra postage.

Explanation of Solution

Given information:

Weight, $x=1$

Mean, $\mu =0.69$

Standard deviation, $\sigma =0.16$

Calculations:

Calculate the Z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{1-0.69}{0.16}\approx 1.94$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with 1.9 and the column that starts with .04 of the standard normal probability table for $P\left(z<-1.79\right)$ .

  $\begin{array}{l}P\left(x>0.1\right)=P\left(z>1.94\right)\\ =1-P\left(z<1.94\right)\\ =1-0.9738\\ =0.0262\end{array}$

Therefore,

Around 0.0262 of the first − class letters at this post office require extra postage.