Given the information about the Acme Company's widget weights and the bell-shaped (normal) distribution, we can apply the Empirical Rule (also known as the 68-95-99.7 rule). This rule states that for a normal distribution:

• 68% of the data lies within 1 standard deviation from the mean.
• 95% of the data lies within 2 standard deviations from the mean.
• 99.7% of the data lies within 3 standard deviations from the mean.

a) 95% of the widget weights lie between:

To find the range that contains 95% of the widget weights, we calculate the values that are 2 standard deviations away from the mean:

• Mean $\mu = 61$ ounces
• Standard deviation $\sigma = 9$ ounces

Lower bound: $61 - 2 \times 9 = 61 - 18 = 43 \text{ ounces}$ Upper bound: $61 + 2 \times 9 = 61 + 18 = 79 \text{ ounces}$ So, 95% of the widget weights lie between 43 ounces and 79 ounces, which is already provided as correct.

b) What percentage of the widget weights lie between 34 and 79 ounces?

To calculate this, we first express these values in terms of standard deviations from the mean:

• $34$ ounces is: $\frac{34 - 61}{9} = \frac{-27}{9} = -3 \sigma$
• $79$ ounces is: $\frac{79 - 61}{9} = \frac{18}{9} = 2 \sigma$

According to the Empirical Rule:

• From $-3 \sigma$ to $+3 \sigma$ contains 99.7% of the data.
• From $-2 \sigma$ to $+2 \sigma$ contains 95% of the data.

From $-3 \sigma$ to $+2 \sigma$ would contain $99.7\% - (100\% - 95\%) = 99.7\% - 5\% = 94.7\%$.

Thus, 94.7% of the widget weights lie between 34 ounces and 79 ounces.

c) What percentage of the widget weights lie below 70 ounces?

First, we calculate how many standard deviations 70 is from the mean:

$\frac{70 - 61}{9} = \frac{9}{9} = 1 \sigma$

According to the Empirical Rule:

• 68% of the data lies between $-1 \sigma$ and $+1 \sigma$, so from 52 to 70 ounces, 34% lies between the mean (61 ounces) and $+1 \sigma$ (70 ounces).

Since the normal distribution is symmetric, 50% of the data lies below the mean (61 ounces), and we add the 34% above the mean:

$50\% + 34\% = 84\%$

Thus, 84% of the widget weights lie below 70 ounces.

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Would you like any further clarification or a more detailed explanation of any part of this?

5 Related Questions:

1. What percentage of the widget weights lie above 79 ounces?
2. What is the probability that a randomly chosen widget weighs less than 43 ounces?
3. What percentage of widget weights lie between 52 and 70 ounces?
4. How many standard deviations is a weight of 88 ounces from the mean?
5. What percentage of widget weights lie between 43 and 70 ounces?

Tip:

When dealing with normal distributions, converting data points to "z-scores" (standard deviations from the mean) simplifies the process of calculating probabilities or percentages.