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?
To answer these questions, we need to use the concept of the standard normal distribution. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. By standardizing our values, we can use a standard normal distribution table or a statistical calculator to find the corresponding probabilities.
Let's go through each question step by step:
(a) To find the percentage of kernels that will fail to pop if the popcorn is cooked for 2 minutes (or 120 seconds), we need to calculate the probability of a kernel taking more than 120 seconds to pop.
First, we standardize the value by subtracting the mean (140 seconds) and dividing by the standard deviation (25 seconds). So, (120 - 140) / 25 = -0.8.
Next, we can use a standard normal distribution table or a statistical calculator to find the probability corresponding to a z-score of -0.8. The table will give us the area to the left of -0.8, but since we want the probability of a kernel taking more than 120 seconds, we need to subtract the area from 1.
The probability of a kernel taking more than 120 seconds to pop is approximately 1 - 0.2119 = 0.7881, or 78.81%.
(b) To find the percentage of kernels that will fail to pop if the popcorn is cooked for 3 minutes (or 180 seconds), we follow the same process.
Standardizing the value, we get (180 - 140) / 25 = 1.6.
Using the standard normal distribution table or a statistical calculator, we find the area to the left of 1.6 is approximately 0.9452.
Since we want the probability of a kernel taking more than 180 seconds to pop, we subtract the area from 1: 1 - 0.9452 = 0.0548, or 5.48%.
(c) If you want 95 percent of the kernels to pop, you need to find the corresponding cooking time.
We want to find the value (time) that corresponds to a probability of 0.95. For this, we need to find the z-score that corresponds to the probability of 0.95 using the standard normal distribution table or a statistical calculator.
The z-score corresponding to 0.95 is approximately 1.645. We can then unstandardize this z-score by multiplying it by the standard deviation (25) and adding the mean (140).
Time = (1.645 * 25) + 140 ≈ 183.625 seconds.
So, if you want 95 percent of the kernels to pop, you would allow approximately 183.625 seconds or 3 minutes and 3.625 seconds of cooking time.
(d) Similarly, if you want 99 percent of the kernels to pop, you find the z-score that corresponds to 0.99 using the standard normal distribution table or a statistical calculator. The z-score is approximately 2.326.
Unstandardizing, we get (2.326 * 25) + 140 = 196.65 seconds.
So, if you want 99 percent of the kernels to pop, you would allow approximately 196.65 seconds or 3 minutes and 16.65 seconds of cooking time.
To solve these questions, we'll use the normal distribution and z-scores.
(a) To find the percentage of kernels that will fail to pop in 2 minutes (120 seconds), we need to find the area under the curve to the left of the z-score corresponding to 120 seconds.
First, we calculate the z-score using the formula:
z = (x - mean) / standard deviation
For a time of 2 minutes (120 seconds):
z = (120 - 140) / 25
z = -0.8
Next, we need to find the area to the left of this z-score. We can use a standard normal distribution table or a calculator to find this area. Looking up the z-score of -0.8 in the table, we find that the area to the left is approximately 0.2119.
So, approximately 21.19% of the kernels will fail to pop if the popcorn is cooked for 2 minutes.
(b) To find the percentage of kernels that will fail to pop in 3 minutes (180 seconds), we follow the same steps as above.
For a time of 3 minutes (180 seconds):
z = (180 - 140) / 25
z = 1.6
Looking up the z-score of 1.6 in the table, we find that the area to the left is approximately 0.9452.
So, approximately 94.52% of the kernels will fail to pop if the popcorn is cooked for 3 minutes.
(c) To find the time needed for 95% of the kernels to pop, we need to find the corresponding z-score that gives us an area of 0.95 to the left.
Using a standard normal distribution table, we can find that the z-score for an area of 0.95 is approximately 1.645.
Now, we can substitute this z-score back into the formula to find the time:
1.645 = (x - 140) / 25
Solving for x:
x - 140 = 1.645 * 25
x - 140 = 41.125
x = 181.125
So, you would allow approximately 181.125 seconds (or about 3 minutes and 1 second) if you wanted 95% of the kernels to pop.
(d) Similarly, to find the time needed for 99% of the kernels to pop, we find the z-score for an area of 0.99 using the standard normal distribution table. The z-score for 0.99 is approximately 2.326.
Using the same formula:
2.326 = (x - 140) / 25
Solving for x:
x - 140 = 2.326 * 25
x - 140 = 58.15
x = 198.15
So, you would allow approximately 198.15 seconds (or about 3 minutes and 18 seconds) if you wanted 99% of the kernels to pop.
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.