Question

Post office A local post office weighs outgoing mail and finds that the weights of first-class letters are approximately Normally distributed with a mean of $0.69$ ounces and a standard deviation of $0.16$ ounces.

a. Estimate the 60th percentile of first-class letter weights.

b. First-class letters weighing more than $1$ ounce require extra postage. What proportion of first-class letters at this post office requires extra postage?

Short answer

Part (a)$60th$percentile is $0.73$ounces.

Part (b) Around $0.0262$of the first $-$class letters at this post office require extra postage.

Step-by-step solution

Part (a) Step 1: Given information

Mean,$\mu =0.69$

Standard deviation,$\sigma =0.16$

Part (a) Step 2: Concept

The formula used:$z=\frac{x-\mu }{\sigma }$

Part (a) Step 3: Calculation

The $Xth$ percentile represents a data value with $x$ percent of the data values below it.

That implies

The $60th$ percentile represents a data value with $60\%$ of the data values below it.

Now,

In the normal probability table of the appendix, find the $z-$score that corresponds to a probability of $60\%$ (or $0.60$).

Note that

The probability that comes closest is $0.5987$ which is in row $0.2$ and column$.05$ of the normal distribution.

probability table.

Then

The corresponding $z-$ score,

$z=0.25$

In order to find $nth$ percentile, 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,

$60th$the percentile will be $0.73$ ounces.

Part (b) Step 1: Calculation

Calculate the Z − score,

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

To find the equivalent probability, use the normal probability table in the appendix.

The usual normal probability table for $P(z<-1.79)$has a row that starts with $1.9$and a column that starts with$.04$

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

Therefore,

Around $0.0262$ of the first$-$ class letters at this post off ice require extra postage.