(a)
Percentage of large lids those are too small to fit.
(a)
Answer to Problem 69E
Around 6.68% of large lids are too small to fit.
Explanation of Solution
Given:
Lid diameter,
Mean,
Standard deviation,
Calculations:
Calculate the z − score,
Use normal
See the row that starts with -1.5 and the column that starts with .00 of the standard normal probability table for
Thus,
Around 6.68% of the large lids are too small to fit.
(b)
Percentage of large lids those are too big to fit.
(b)
Answer to Problem 69E
Around 0.02% of the large lids are too big to fit.
Explanation of Solution
Given:
Lid diameter,
Mean,
Standard deviation,
Calculations:
Calculate the Z − score,
Use normal probability table in the appendix, to find the corresponding probability.
3.50 do not exist in the table, but 3.49 exist.
See the row that starts with 3.5 and the column that starts with .00 of the standard normal probability table for
Thus,
Around 0.02% of the large lids are too big to fit.
(c)
Whether it makes any sense for the lid manufacturer to try to make one of these values larger than the other.
(c)
Answer to Problem 69E
It does not make any sense for the lid manufacturer to try to make one of these values larger than the other.
Explanation of Solution
Given:
Lid diameter,
Mean,
Standard deviation,
Calculations:
According to Part (a) result,
Around 6.68% of the large lids are too small to fit.
According to Part (b) result,
Around 0.02% of the large lids are too big to fit.
Note that
There are much more lids that are very small compared to the lids that will be very large.
Hence,
It does not make any sense to try to make one of these values larger than the other, due to the lids won’t fit in both cases (they won’t fit when they are small and they won’t fit when they are too large).
Chapter 2 Solutions
PRACTICE OF STATISTICS F/AP EXAM
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