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

(a)

To determine

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

(a)

Expert Solution
Check Mark

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=3.953.980.02=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
Check Mark

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=4.053.980.02=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
Check Mark

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 Solutions

The Practice of Statistics for AP - 4th Edition

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