91影视

Lid diameter,$x=3.95$

Mean,$渭=3.98$

Standard deviation,$蟽=0.02$

The formula used:$z=\frac{x鈭�渭}{蟽}$

Calculate the $z鈭�$score,

$z=\frac{x鈭�渭}{蟽}=\frac{3.95鈭�3.98}{0.02}=鈭�1.50$

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

See the usual normal probability table for $P(z<鈭�1.50)$and the row that starts with $-1.5$and the column that starts with$.00$

$$\begin{gathered} P(x<3.95)=P(z<鈭�1.50) \\ =0.0668 \\ =6.68\% \end{gathered}$$

Thus,

Approximately $6.68$ percent of large lids will not fit.

Calculate the $Z鈭�$score,

$z=\frac{x鈭�渭}{蟽}=\frac{4.05鈭�3.98}{0.02}=3.50$

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

$3.50$do not exist in the table, but $3.49$exist.

See the typical normal probability table for $P(z<3.50)$and the row that starts with $3.5$and the column that starts with$.00$

$$\begin{gathered} P(x>4.05)=P(z>3.50) \\ =1鈭�P(z<3.50)鈮�1鈭�P(z<3.49) \\ =1鈭�0.9998 \\ =0.0002 \\ =0.02\% \end{gathered}$$

Thus

Approximately $0.02\%$ of the large lids are too big to fit.

According to Part (a) result,

Approximately $6.68$ percent of large lids will not fit.

According to Part $\left(b\right)$result,

Approximately$0.02\%$ of the large lids are too big to fit.

Note that

In comparison to the lids that will be very large, there are a lot more little lids.

Hence,

It's pointless to try to make one of these values larger than the other because the lids will not fit in either scenario (they won't fit when tooled tiny, and they won't fit when tooled large).