The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 57 ounces and a standard deviation of 6 ounces. Use the Empirical Rule and a sketch of the normal distribution in order to answer these questions. a) $$ 95 \% $$ of the widget weights lie between b) What percentage of the widget weights lie between 39 and 69 ounces? c) What percentage of the widget weights lie above 51 ?

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Let's analyze the widget weights manufactured by Acme Company using the **Empirical Rule**, which is applicable since the distribution is bell-shaped (normal distribution). The Empirical Rule states:

- **68%** of the data falls within **±1 standard deviation** of the mean.
- **95%** of the data falls within **±2 standard deviations** of the mean.
- **99.7%** of the data falls within **±3 standard deviations** of the mean.

Given:
- **Mean ($$ \mu $$)** = 57 ounces
- **Standard Deviation ($$ \sigma $$)** = 6 ounces

### a) $$ 95\% $$ of the widget weights lie between

According to the **Empirical Rule**, **95%** of the data lies within **±2 standard deviations** from the mean.

$$
\begin{align*}
\text{Lower Bound} &= \mu - 2\sigma = 57 - 2(6) = 57 - 12 = 45 \text{ ounces} \
\text{Upper Bound} &= \mu + 2\sigma = 57 + 2(6) = 57 + 12 = 69 \text{ ounces}
\end{align*}
$$

**Answer:**  
$$ 95\% $$ of the widget weights lie between **45 ounces** and **69 ounces**.

---

### b) What percentage of the widget weights lie between 39 and 69 ounces?

First, determine how many standard deviations away 39 and 69 ounces are from the mean.

$$
\begin{align*}
\text{For 39 ounces:} \quad z &= \frac{39 - 57}{6} = \frac{-18}{6} = -3 \
\text{For 69 ounces:} \quad z &= \frac{69 - 57}{6} = \frac{12}{6} = 2
\end{align*}
$$

So, 39 ounces is **$$ \mu - 3\sigma $$** and 69 ounces is **$$ \mu + 2\sigma $$**.

Using the Empirical Rule:
- **$$ \mu - 3\sigma $$ to $$ \mu + 3\sigma $$** covers **99.7%** of the data.
- **$$ \mu - 2\sigma $$ to $$ \mu + 2\sigma $$** covers **95%** of the data.

To find the percentage between **39 ounces ($$ \mu - 3\sigma $$)** and **69 ounces ($$ \mu + 2\sigma $$)**:
1. **From $$ \mu - 3\sigma $$ to $$ \mu - 2\sigma $$:**  
   This range represents the lower tail beyond $$ \mu - 2\sigma $$, which is **2.5%**.

2. **From $$ \mu - 2\sigma $$ to $$ \mu + 2\sigma $$:**  
   This is **95%** of the data.

3. **From $$ \mu + 2\sigma $$ to $$ \mu + 3\sigma $$:**  
   Not included in this range, so we ignore the upper tail beyond $$ \mu + 2\sigma $$.

Adding these together:
$$
2.5\% (\text{lower tail}) + 95\% (\text{central range}) = 97.5\%
$$

**Answer:**  
Approximately **97.5%** of the widget weights lie between **39 ounces** and **69 ounces**.

---

### c) What percentage of the widget weights lie above 51 ounces?

First, determine how many standard deviations away 51 ounces is from the mean.

$$
z = \frac{51 - 57}{6} = \frac{-6}{6} = -1
$$

So, 51 ounces is **$$ \mu - 1\sigma $$**.

Using the Empirical Rule:
- **$$ \mu - 1\sigma $$ to $$ \mu + 1\sigma $$** covers **68%** of the data.
- The remaining **32%** is split equally between the two tails, with **16%** in each tail.

To find the percentage above 51 ounces ($$ \mu - 1\sigma $$):
1. **From $$ \mu - 1\sigma $$ to $$ \mu + 1\sigma $$:**  
   This range covers **68%** of the data.

2. **Above $$ \mu + 1\sigma $$:**  
   This accounts for **16%** of the data.

Adding these together:
$$
68\% (\text{central range}) + 16\% (\text{upper tail}) = 84\%
$$

**Answer:**  
Approximately **84%** of the widget weights lie above **51 ounces**.
a) $$ 95\% $$ of the widget weights lie between **45 ounces** and **69 ounces**. b) Approximately **97.5%** of the widget weights lie between **39 ounces** and **69 ounces**. c) Approximately **84%** of the widget weights lie above **51 ounces**.

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