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At some fast-food restaurants, customers who want a lid for their drinks get them from a large stack near the straws, napkins, and condiments. The lids are made with a small amount of flexibility so they can be stretched across the mouth of the cup and then snugly secured. When lids are too small or too large, customers can get very frustrated, especially if they end up spilling their drinks. At one particular restaurant, large drink cups require lids with a 鈥渄iameter鈥� of between $3.95$ and $4.05$ inches. The restaurant鈥檚 lid supplier claims that the diameter of its large lids follows a Normal distribution with a mean of $3.98$ inches and a standard deviation of $0.02$ inches. Assume that the supplier鈥檚 claim is true.

Put a lid on it! The supplier is considering two changes to reduce to $1\%$ the percentage of its large-cup lids that are too small. One strategy is to adjust the mean diameter of its lids. Another option is to alter the production process, thereby decreasing the standard deviation of the lid diameters.

a. If the standard deviation remains at $蟽=0.02$ inch, at what value should the

supplier set the mean diameter of its large-cup lids so that only $1\%$ is too small to fit?

b. If the mean diameter stays at $渭=3.98$ inches, what value of the standard

deviation will result in only $1\%$ of lids that are too small to fit?

c. Which of the two options in parts (a) and (b) do you think is preferable? Justify your answer. (Be sure to consider the effect of these changes on the percent of lids that are too large to fit.)$蟽=0.02$

Step by step solution

01

Part (a) Step 1: Given information

Lid diameter, $x=3.95$

Standard deviation,$蟽=0.02$

02

Part (a) Step 2: Concept

The formula used:$z=\frac{x鈭�渭}{蟽}$

03

Part (a) Step 3: Calculation

The probability to the left of the $z-$score, or for values smaller than the $z-$ score, are listed in Table $A$

We know

$1\%$is less than $z$

The value of the $z-$score in Table corresponds to a chance of $0.01$(or $1\%$). $鈭�A$:

$z=鈭�2.33$

Calculate the $z-$ score:

$z=\frac{x鈭�渭}{蟽}$

Multiply both sides by $蟽$:

$z蟽=x鈭�渭$

Add $渭$to both sides:

$渭+z蟽=x$

Subtract $z蟽$from both sides:

$渭=x鈭�z蟽$

Substitute values and calculate:

$渭=x鈭�z蟽=3.95鈭�(鈭�2.33)(0.02)=3.9966$

04

Part (b) Step 1: Calculation

The probability to the left of the $z-$score, or for values smaller than the $z-$score, are listed in Table $A$

We know

$1\%$is less than $z$

Corresponding to the probability of $0.01$(or $1\%$), the value of $z鈭�$score in Table $鈭�A$:

$z=鈭�2.33$

Calculate the $z鈭�$score:

$z=\frac{x鈭�渭}{蟽}$

Multiply both sides by $蟽$:

$z蟽=x鈭�渭$

Subtract $z$ from both sides:

$蟽=\frac{x鈭�渭}{z}$

Calculate by substituting values:

$蟽=\frac{3.95鈭�3.98}{鈭�2.33}=0.0129$

05

Part (c) Step 1: Given information

Result from Part (a),

Mean,$渭=3.9966$

Standard deviation,$蟽=0.02$

Result from Part (b),

Mean,$渭=3.98$

Standard deviation,$蟽=0.0129$

06

Part (c) Step 2: Explanation

The lids are too big if their diameter exceeds $4.05$inches.

Calculate the $z鈭�$score:

$z=\frac{x鈭�渭}{蟽}=\frac{4.05鈭�3.9966}{0.02}=2.67$

Or

$z=\frac{x鈭�渭}{蟽}=\frac{4.05鈭�3.98}{0.0129}=5.43$

To find the equivalent probability, use Table $A$:

$$\begin{gathered} P(x>4.05)=P(z>2.67) \\ =1鈭�P(z<2.67) \\ =1鈭�0.9962 \\ =0.0038 \\ =0.38\% \end{gathered}$$

Or

$$\begin{gathered} P(x>4.05)=P(z>5.43) \\ =1鈭�P(z<5.43) \\ =1鈭�0.9999 \\ =0.0001 \\ =0.01\% \end{gathered}$$

We need to fit as many lids as feasible.

Thus,

We'll go with the lower of the two percentages calculated above.

Therefore,

The probability of $0.01$percent, which corresponds to Part $1$will be preferred $\left(b\right)$

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