Question

competency 14 (2 of 2): quiz

1. the amount of soda dispensed in a paper cup of an automatic machine has a mean of 6.2 ounces and a standard deviation of 0.4 ounces. what percent of all the paper cups will have between 5 and 7 ounces of soda?

a. 6.5%
b. 4.2%
c. 2.7%
d. 0.3%

Explanation:

⚡ Using what you learned: normal distribution

Step 1: Calculate z-scores

Calculate the \(z\)-scores for the boundaries \(x_1 = 5\) and \(x_2 = 7\) using the mean \(\mu = 6.2\) and standard deviation \(\sigma = 0.4\):

$$ z_1 = \frac{5 - 6.2}{0.4} = \frac{-1.2}{0.4} = -3 $$

$$ z_2 = \frac{7 - 6.2}{0.4} = \frac{0.8}{0.4} = 2 $$

Step 2: Find the area between z-scores

Using the Empirical Rule (68-95-99.7 rule):

• The area within 2 standard deviations of the mean (\(z = -2\) to \(z = 2\)) is approximately \(95\%\).
• The area within 3 standard deviations of the mean (\(z = -3\) to \(z = 3\)) is approximately \(99.7\%\).

To find the area between \(z = -3\) and \(z = 2\):

$$ P(-3 < Z < 2) = P(-3 < Z < 0) + P(0 < Z < 2) $$

Using the symmetry of the normal distribution:

• \(P(-3 < Z < 0) = \frac{0.997}{2} = 0.4985\) (or \(49.85\%\))
• \(P(0 < Z < 2) = \frac{0.95}{2} = 0.475\) (or \(47.5\%\))

Sum these two regions:

$$ P(-3 < Z < 2) = 0.4985 + 0.475 = 0.9735 \text{ (or } 97.35\%\text{)} $$

Step 3: Calculate the remaining percentage

The question asks for the percentage of paper cups that will have between 5 and 7 ounces of soda. Since \(97.35\%\) of the cups fall inside this range, we find the percentage that falls outside this range to match the options, or we look for the closest representation.

Let's re-read the question: "What percent of all the paper cups will have between 5 and 7 ounces of soda."
Usually, this is a typo in the test question, and it meant to ask "What percent of all the paper cups will not have between 5 and 7 ounces of soda" or "will have outside 5 and 7 ounces of soda."

Let's calculate the percentage outside this interval:

$$ 100\% - 97.35\% = 2.65\% \approx 2.7\% $$

This matches option C.

Answer:

C. 2.7%