1 answer

In a certain microwave oven on the high power setting, the time it takes a randomly chosen kernel of popcorn

Question:

In a certain microwave oven on the high power setting, the time it takes a randomly chosen kernel
of popcorn to pop is normally distributed with a mean of 140 seconds and a standard deviation of
25 seconds. What percentage of the kernels will fail to pop if the popcorn is cooked for (a) 2 minutes?
(b) Three minutes? (c) If you wanted 95 percent of the kernels to pop, what time would you
allow? (d) If you wanted 99 percent to pop?

Answers

You will need to use the z-score formula and a z-table to answer these questions.

The formula:
z = (x - mean)/sd -->sd = standard deviation

For a) and b), find z using the formula. Then determine the percentage using the z-table.
For c) and d), find x using the formula (which means you will need to find z equating to 95% and 99% in the table).

I'll get you started and let you take it from there.
a) z = (120 - 140)/25
b) z = (180 - 140)/25
c) 1.28 = (x - 140)/25
d) 2.33 = (x - 140)/25

I hope this will help..

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