Posted by **Andrew** on Monday, May 12, 2008 at 9:38pm.

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?

- Statistics -
**MathGuru**, Thursday, May 15, 2008 at 8:08am
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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