(a) Answer Approximately 6.7% of lids are too small. Explanation Given: Mean diameter is 3.98 inches Standard deviation is 0.02 inches. Diameter is between 3.95 and 4.05 inches Calculation: If a lid is too small to fit, that means that its diameter is less than 3.95. Let x be the width of lid. Find the proportion of lids for which x < 3.95 . The graph below shows the area below 3.95 Find the z value corresponding to x = 3.95 as follows: z = 3.95 − 3.98 0.02 = − 1.5 From standard normal tables, the proportion of observations less than − 1.5 is 0.0668. So, approximately 6.7% of lids are too small. Conclusion: Therefore, 6.7% of lids are too small. (b) To determine To find: the percent of large lids that are very big to fit. Answer It is extremely rare for a lid to be too big. Explanation Given: Mean diameter is 3.98 inches Standard deviation is 0.02 inches. Diameter is between 3.95 and 4.05 inches Calculation: If a lid is too large to fit, that means that its diameter is more than 4.05. Let x be the width of lid. Find the proportion of lids for which x > 4.05 . The graph below shows the area above 4.05: Find the z value corresponding to 4.05 as follows: z = 4.05 − 3.98 0.02 = 3.5 From the standard normal tables, the proportion of observations less than 3.5 is approximately 1, so the proportion of observations greater than 3.5 is approximately 0. Hence, it is extremely rare for a lid to be too big. Conclusion: Hence, it is extremely rare for a lid to be too big. (c) To determine To compare: the answers to (a) and (b). Answer It makes more sense to have a larger proportion of lids too small rather than too large. Explanation Given: Mean diameter is 3.98 inches Standard deviation is 0.02 inches. Diameter is between 3.95 and 4.05 inches Calculation: It makes more sense to have a larger proportion of lids too small rather than too large. The company wants to make sure that the fit is snug. If more lids are too large, there will be more spills. If lids are too small, customers will just try another lid. But if lids are too large, the customer may not notice and then spill the drink. Conclusion: Therefore, it makes more sense to have a larger proportion of lids too small rather than too large.

The Practice of Statistics for AP - 4th Edition

4th Edition

ISBN: 9781429245593

Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.

Publisher: Macmillan Higher Education

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Question

Chapter 2.2, Problem 56E

(a)

To determine

To find: the percent of large lids that are very small to fit.

(a)

Expert Solution

Answer to Problem 56E

Approximately 6.7% of lids are too small.

Explanation of Solution

Given:

Mean diameter is 3.98 inches

Standard deviation is 0.02 inches.

Diameter is between 3.95 and 4.05 inches

Calculation:

If a lid is too small to fit, that means that its diameter is less than 3.95. Let x be the width of lid.

Find the proportion of lids for which x<3.95 .

The graph below shows the area below 3.95

The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 56E , additional homework tip 1

Find the z value corresponding to x=3.95 as follows:

z=3.95−3.980.02=−1.5

From standard normal tables, the proportion of observations less than −1.5 is 0.0668.

So, approximately 6.7% of lids are too small.

Conclusion:

Therefore, 6.7% of lids are too small.

(b)

To determine

To find: the percent of large lids that are very big to fit.

(b)

Expert Solution

Answer to Problem 56E

It is extremely rare for a lid to be too big.

Explanation of Solution

Given:

Mean diameter is 3.98 inches

Standard deviation is 0.02 inches.

Diameter is between 3.95 and 4.05 inches

Calculation:

If a lid is too large to fit, that means that its diameter is more than 4.05. Let x be the width of lid.

Find the proportion of lids for which x>4.05 .

The graph below shows the area above 4.05:

The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 56E , additional homework tip 2

Find the z value corresponding to 4.05 as follows:

z=4.05−3.980.02=3.5

From the standard normal tables, the proportion of observations less than 3.5 is approximately 1, so the proportion of observations greater than 3.5 is approximately 0.

Hence, it is extremely rare for a lid to be too big.

Conclusion:

Hence, it is extremely rare for a lid to be too big.

(c)

To determine

To compare: the answers to (a) and (b).

(c)

Expert Solution

Answer to Problem 56E

It makes more sense to have a larger proportion of lids too small rather than too large.

Explanation of Solution

Given:

Mean diameter is 3.98 inches

Standard deviation is 0.02 inches.

Diameter is between 3.95 and 4.05 inches

Calculation:

It makes more sense to have a larger proportion of lids too small rather than too large. The company wants to make sure that the fit is snug. If more lids are too large, there will be more spills. If lids are too small, customers will just try another lid. But if lids are too large, the customer may not notice and then spill the drink.

Conclusion:

Therefore, it makes more sense to have a larger proportion of lids too small rather than too large.

Chapter 2.2, Problem 55E

Chapter 2.2, Problem 57E

Chapter 2 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 2.1 - Prob. 1.1CYU Ch. 2.1 - Prob. 1.2CYU Ch. 2.1 - Prob. 1.3CYU Ch. 2.1 - Prob. 1.4CYU Ch. 2.1 - Prob. 2.1CYU Ch. 2.1 - Prob. 2.2CYU Ch. 2.1 - Prob. 2.3CYU Ch. 2.1 - Prob. 3.1CYU Ch. 2.1 - Prob. 3.2CYU Ch. 2.1 - Prob. 3.3CYU

Ch. 2.1 - Prob. 4.1CYU Ch. 2.1 - Prob. 4.2CYU Ch. 2.1 - Prob. 4.3CYU Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.2 - Prob. 1.1CYU Ch. 2.2 - Prob. 1.2CYU Ch. 2.2 - Prob. 1.3CYU Ch. 2.2 - Prob. 2.1CYU Ch. 2.2 - Prob. 2.2CYU Ch. 2.2 - Prob. 2.3CYU Ch. 2.2 - Prob. 2.4CYU Ch. 2.2 - Prob. 2.5CYU Ch. 2.2 - Prob. 3.1CYU Ch. 2.2 - Prob. 3.2CYU Ch. 2.2 - Prob. 3.3CYU Ch. 2.2 - Prob. 41E Ch. 2.2 - Prob. 42E Ch. 2.2 - Prob. 43E Ch. 2.2 - Prob. 44E Ch. 2.2 - Prob. 45E Ch. 2.2 - Prob. 46E Ch. 2.2 - Prob. 47E Ch. 2.2 - Prob. 48E Ch. 2.2 - Prob. 49E Ch. 2.2 - Prob. 50E Ch. 2.2 - Prob. 51E Ch. 2.2 - Prob. 52E Ch. 2.2 - Prob. 53E Ch. 2.2 - Prob. 54E Ch. 2.2 - Prob. 55E Ch. 2.2 - Prob. 56E Ch. 2.2 - Prob. 57E Ch. 2.2 - Prob. 58E Ch. 2.2 - Prob. 59E Ch. 2.2 - Prob. 60E Ch. 2.2 - Prob. 61E Ch. 2.2 - Prob. 62E Ch. 2.2 - Prob. 63E Ch. 2.2 - Prob. 64E Ch. 2.2 - Prob. 65E Ch. 2.2 - Prob. 66E Ch. 2.2 - Prob. 67E Ch. 2.2 - Prob. 68E Ch. 2.2 - Prob. 69E Ch. 2.2 - Prob. 70E Ch. 2.2 - Prob. 71E Ch. 2.2 - Prob. 72E Ch. 2.2 - Prob. 73E Ch. 2.2 - Prob. 74E Ch. 2.2 - Prob. 75E Ch. 2.2 - Prob. 76E Ch. 2 - Prob. 1CRE Ch. 2 - Prob. 2CRE Ch. 2 - Prob. 3CRE Ch. 2 - Prob. 4CRE Ch. 2 - Prob. 5CRE Ch. 2 - Prob. 6CRE Ch. 2 - Prob. 7CRE Ch. 2 - Prob. 8CRE Ch. 2 - Prob. 9CRE Ch. 2 - Prob. 10CRE Ch. 2 - Prob. 11CRE Ch. 2 - Prob. 12CRE Ch. 2 - Prob. 1PT Ch. 2 - Prob. 2PT Ch. 2 - Prob. 3PT Ch. 2 - Prob. 4PT Ch. 2 - Prob. 5PT Ch. 2 - Prob. 6PT Ch. 2 - Prob. 7PT Ch. 2 - Prob. 8PT Ch. 2 - Prob. 9PT Ch. 2 - Prob. 10PT Ch. 2 - Prob. 11PT Ch. 2 - Prob. 12PT Ch. 2 - Prob. 13PT

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