Chapter 2.2, Problem 70E

(a)

To determine

Value at which supplier should set the mean diameter of its large cup lids so that only 1% are too small to fit.

Answer to Problem 70E

Mean diameter value,

  $\mu =3.9966$

Explanation of Solution

Given:

Lid diameter, $x=3.95$

Standard deviation, $\sigma =0.02$

Calculations:

Table − A consists of the probabilities to the left of the z − score or for the values smaller than the z − score.

We know

1% is less than z .

Corresponding to the probability of 0.01 (or 1%), the value of z − score in Table − A:

  $z=-2.33$

Calculate the z − score:

  $z=\frac{x-\mu }{\sigma }$

Multiply both sides by $\sigma$ :

  $z\sigma =x-\mu$

Add $\mu$ to both sides:

  $\mu +z\sigma =x$

Subtract $z\sigma$ from both sides:

  $\mu =x-z\sigma$

Substitute values and calculate:

  $\mu =x-z\sigma =3.95-\left(-2.33\right)\left(0.02\right)=3.9966$

(b)

To determine

Value of the standard deviation that result in only 1% of lids that are too small to fit.

Answer to Problem 70E

Standard deviation value,

  $\sigma =0.0129$

Explanation of Solution

Given:

Lid diameter, $x=3.95$

Mean, $\mu =3.98$

Calculations:

Table − A consists of the probabilities to the left of the z − score or for the values smaller than the z − score.

We know

1% is less than z .

Corresponding to the probability of 0.01 (or 1%), the value of z − score in Table − A:

  $z=-2.33$

Calculate the z − score:

  $z=\frac{x-\mu }{\sigma }$

Multiply both sides by $\sigma$ :

  $z\sigma =x-\mu$

Divide both sides by z :

  $\sigma =\frac{x-\mu }{z}$

Substitute values and calculate:

  $\sigma =\frac{3.95-3.98}{-2.33}=0.0129$

(c)

To determine

Preferable option between Part (a) and Part (b).

Answer to Problem 70E

Part (b) will be preferred.

Explanation of Solution

Given:

Result from Part (a),

Mean, $\mu =3.9966$

Standard deviation, $\sigma =0.02$

Result from Part (b),

Mean, $\mu =3.98$

Standard deviation, $\sigma =0.0129$

Calculations:

If the diameter exceeds 4.05, the lids are too large.

Calculate the z − score:

  $z=\frac{x-\mu }{\sigma }=\frac{4.05-3.9966}{0.02}=2.67$

Or

  $z=\frac{x-\mu }{\sigma }=\frac{4.05-3.98}{0.0129}=5.43$

Use Table − A to find the corresponding probability:

  $\begin{array}{l}P\left(x>4.05\right)=P\left(z>2.67\right)\\ =1-P\left(z<2.67\right)\\ =1-0.9962\\ =0.0038\\ =0.38\%\end{array}$

Or

  $\begin{array}{l}P\left(x>4.05\right)=P\left(z>5.43\right)\\ =1-P\left(z<5.43\right)\\ =1-0.9999\\ =0.0001\\ =0.01\%\end{array}$

We require as many lids as possible to fit.

Thus,

We will opt for the smaller percentage from the two calculated above.

Therefore,

The probability of 0.01% will be preferred which corresponds with Part (b).