Question

The Foldgers Black Silk K-cup coffee pods weigh on average .28 ounces with a standard deviation of .05 ounces. Answer each of the questions.
68%, 2.5%, 81.5%, 95%, 41-42, 20-21, 16%, 84%, 121-122, 5%, 87-88, 64, 35-36
What percentage of K-cups can be expected to fall within 2 Standard Deviations of the mean?

What is the probability that a K-cup will have more than .33 ounces?

Sam's club sells K-cups in a box of 128. How many of them will weigh less than .23 ounces?

K-cups that are more than 2 SD below the mean are considered to be to "weak". What percentage of K-cup are considered to "weak"

In a standard box of 44 k-cups, how many will weigh between .23 and .38 ounces

The ideal weight for K-cups are within 1 SD of the mean, how many of the Sam's Club box of K-cups are ideal?

Answer 1

Using the Empirical Rule, it is found that:

• 95% of K-cups can be expected to fall within 2 Standard Deviations of the mean.
• 0.16 = 16% probability that a K-cup will have more than .33 ounces.
• 20 of them will weight less than 0.23 ounces.
• 2.5% of K-cups are considered weak.
• 36 will weigh between 0.23 and 0.38 ounces.
• 30 of the Sam's Club box of K-cups are ideal.

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem:

• The mean is of 0.28 ounces.
• The standard deviation is of 0.05 ounces.

First, by the Empirical Rule, 95% of K-cups can be expected to fall within 2 Standard Deviations of the mean.

0.33 ounces is one standard deviation above the mean.

• 68% of the measures are between 0.23 and 0.33 ounces.
• The normal distribution is symmetric, which means that of the remaining 100 - 68 = 32%, 16% are less than 0.23 ounces and 16% are more than 0.33 ounces, thus:

0.16 = 16% probability that a K-cup will have more than .33 ounces.

16% weigh less than 0.23 ounces. Out of 128:

$0.16(128) = 20$

20 of them will weight less than 0.23 ounces.

5% of the measures are more than 2 SD from the mean. Due to the symmetry of the normal distribution, 2.5% are more than 2 SD below the mean, thus, 2.5% of K-cups are considered weak.

• 0.23 is one standard deviation below the mean.
• 0.38 is two standard deviations above the mean.
• Of the 50% below the mean, 68% are above 0.23.
• Of the 50% above the mean, 95% are below 0.38.

Thus, the proportion between 0.23 and 0.38 is:

$p = 0.68(0.50) + 0.95(0.5) = 0.815$

Out of 44:

$0.815(44) = 36$

36 will weigh between 0.23 and 0.38 ounces.

68% are within 1 standard deviation of the mean.

Out of 44:

$0.68(44) = 30$

30 of the Sam's Club box of K-cups are ideal.

A similar problem is given at brainly.com/question/13503878