Let's analyze this problem.

Given information:

• The mean weight of the widgets is $\mu = 61$ ounces.
• The standard deviation is $\sigma = 5$ ounces.
• The distribution of widget weights is bell-shaped, meaning it's normally distributed.

Part (a):

We are asked to determine the range for which 99.7% of the widget weights lie.

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

• 99.7% of the data lies within 3 standard deviations from the mean in a normal distribution.

So, the range will be: $\text{Lower bound} = \mu - 3\sigma = 61 - 3(5) = 61 - 15 = 46 \, \text{ounces}$ $\text{Upper bound} = \mu + 3\sigma = 61 + 3(5) = 61 + 15 = 76 \, \text{ounces}$

Thus, 99.7% of the weights lie between 46 and 76 ounces.

Part (b):

We are asked what percentage of widget weights lie between 51 and 76 ounces.

To solve this, let's use the standard normal distribution:

1. Convert the weights to z-scores using the formula: $z = \frac{x - \mu}{\sigma}$ For $x = 51$: $z = \frac{51 - 61}{5} = \frac{-10}{5} = -2$ For $x = 76$: $z = \frac{76 - 61}{5} = \frac{15}{5} = 3$

2. Now, using a standard normal distribution table or calculator:

• The area under the curve for $z = -2$ is approximately 0.0228 (2.28% below 51).
• The area under the curve for $z = 3$ is approximately 0.9987 (99.87% below 76).

3. The percentage of widget weights between 51 and 76 ounces is: $0.9987 - 0.0228 = 0.9759 \, \text{or} \, 97.59\%$

Thus, 97.59% of the widget weights lie between 51 and 76 ounces.

Would you like further details or explanations on any part of this?

Here are some related questions:

1. How does the Empirical Rule work for 68% and 95% intervals?
2. What is the formula for finding a z-score?
3. How can I calculate percentages for values outside the 3-standard deviation range?
4. How does standard deviation affect the spread of data in a normal distribution?
5. Can the normal distribution be used for skewed data sets?

Tip: When using the Empirical Rule, remember that it applies specifically to normal distributions, where the data is symmetrically distributed around the mean.