91影视

$x=1ounceand1.2ounces$

Mean,$渭=1.05ounces$

Standard deviation,$蟽=0.08ounce$

The formula used:$z=\frac{x鈭�渭}{蟽}$

Calculate the $z鈭�$score,

$z=\frac{x鈭�渭}{蟽}=\frac{1-1.05}{0.08}鈮�鈭�0.63$

And

$z=\frac{x鈭�渭}{蟽}=\frac{1.2-1.05}{0.08}鈮�1.88$

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

$$\begin{gathered} P(1<x<1.2)=P(鈭�0.63<z<1.88)=P(z<1.88)鈭�P(z<鈭�0.63) \\ =0.9699鈭�0.2643 \\ =0.7056 \\ =70.56\% \end{gathered}$$

Therefore,

Between $1$ and $1.2$ ounces of the ketchup has been put on a burger data-custom-editor="chemistry" $70.56\%$ of time.

We have

Between $1$and $1.2$ounces of ketchup is required for at least $99$percent of the burgers.

$P(1<x<1.2)=99\%=0.99$

Note that

Mean is exactly in the middle i.e. $1.1$

Thus,

$99\%$corresponds with the middle $99\%$

Thus,

$\frac{100\%鈭�99\%}{2}=0.5\%$

There are about $0.05\%$burgers on either side of the interval.

This implies

There are $0.5\%$of the burgers with less than $1$ounce of ketchup.

However,

$99\%+0.5\%=99.5\%$

$99.5$percent of the burgers have less than $1.2$ounces of ketchup on them.

The z scores in the normal probability table correspond to a probability of $0.005(0.5$percent) to $0.995(99.5\%)$or the probability nearest.

$z=卤2.58$

Now,

$z鈭�$score:

$z=\frac{x鈭�渭}{蟽}$

Multiply both sides by $蟽$:

Divide each side by $z鈭�$score $(z)$:

$蟽=\frac{x鈭�渭}{z}$

Substitute the following values for the known values:

Note that

$z=鈭�2.58$corresponds with $x=1$

And

$z=2.5$corresponds with $x=1.2$

Then

$蟽=\frac{x鈭�渭}{z}=\frac{1鈭�1.1}{鈭�2.58}=\frac{鈭�0.1}{鈭�2.58}=\frac{0.1}{2.58}鈮�0.03876$

The standard deviation is around $0.03876$ ounces.

The original standard deviation is $0.08$ ounces.

We get $0.04124$ ounces by subtracting the estimated standard deviation from the original standard deviation.

Thus,

We need to reduce the standard deviation by $0.04124$ ounces.