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?

1 answer

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.