Chapter 6.2, Problem 48E

(a)

To determine

Mean of the excess amount of cereal in a randomly selected box.

Answer to Problem 48E

Mean of Y is 1.9845 grams.

Explanation of Solution

Given information:

X : amount of cereal in a randomly selected box measured in grams

Y : excessamount of cereal in a randomly selected box measured in grams

For X :

Mean,

  ${\mu }_X=1.70\text{grams}$

Standard deviation,

  ${\sigma }_X=0.03\text{grams}$

Recall that

  $1\text{ounce}=28.35\text{grams}$

Calculations:

We know that

The advertised amount of cereal is 1.63 ounces.

Then

The excess amount of cereal measured in grams is the amount of cereal X decreased by 1.63.

Now,

Express the amount in grams instead of ounces.

We also know

  $1\text{ounce}=28.35\text{grams}$

Then

Multiply the excess amount by 28.35.

Thus,

  $Y=28.35\left(X-1.63\right)$

Now,

Every data value in the distribution of X is first subtracted by the same constant 1.63 and then multipliedby the same constant 28.35.

If every data value is subtracted by the same constant, the center of the distribution is also subtracted by the same constant.

Also,

If every data value is multiplied by the same constant, the center of the distribution is also multiplied by the same constant.

We know that

The mean is the measure of the center.

Thus,

The mean is first decreased by 1.63 and then multiplied by 28.35.

  ${\mu }_Y=28.35\left({\mu }_X-1.63\right)=28.35\left(1.70-1.63\right)=28.35\left(0.07\right)=1.9845\text{grams}$

(b)

To determine

Interpretation and calculation for standard deviation of Y .

Answer to Problem 48E

Standard deviation of Y ,

  ${\sigma }_Y=0.8505\text{grams}$

Explanation of Solution

Given information:

X : amount of cereal in a randomly selected box measured in grams

Y : excess amount of cereal in a randomly selected box measured in grams

For X :

Mean,

  ${\mu }_X=1.70\text{grams}$

Standard deviation,

  ${\sigma }_X=0.03\text{grams}$

Recall that

  $1\text{ounce}=28.3\text{grams}$

Calculations:

We know that

The advertised amount of cereal is 1.63 ounces.

Then

The excess amount of cereal measured in grams is the amount of cereal X decreased by 1.63.

Now,

Express the amount in grams instead of ounces.

We also know

  $1\text{ounce}=28.3\text{grams}$

Then

Multiply the excess amount 28.35.

Thus,

  $Y=28.35\left(X-1.63\right)$

Now,

Every data value in the distribution of X is first subtracted by the same constant 1.63 and then multiplied by the same constant 28.35.

If every data value is subtracted by the same constant, the spread of the distribution is unaffected.

Also,

If every data value is multiplied by the same constant, the spread of the distribution is also multiplied by the same constant.

We know that

The standard deviation is the measure of the spread.

Thus,

The standard deviation is multiplied by 28.35.

  ${\sigma }_Y=28.35{\sigma }_X=28.35\left(0.03\right)=0.8505\text{grams}$

Thus,

The excess amount of cereal varies on an average by 0.8505 grams from the mean excess amount of 1.9845 grams.

(c)

To determine

Probability of getting at least 1 gram more cereal than advertised.

Answer to Problem 48E

Probability of getting at least 1 gram more cereal is 0.8770.

Explanation of Solution

Given information:

Amount of more cereal,

  $x=1\text{gram}$

From Part (a),

  ${\mu }_Y=1.9845\text{grams}$

From Part (b),

  ${\sigma }_Y=0.8505\text{grams}$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{1-1.9845}{0.8505}\approx -1.16$

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

See the row that starts with -1.1 and the column that starts with .06 of the standard normal probability table for $P\left(z<-1.16\right)$ .

  $\begin{array}{l}P\left(x\ge 1\right)=P\left(z>-1.16\right)\\ =1-P\left(z<-1.16\right)\\ =1-0.1230\\ =0.8770\end{array}$

Thus,

Probability of getting at least 1 gram more cereal than advertised is 0.8770.