Question

The weights of boxes of cereal are normally distributed with a mean of 14 ounces and a standard deviation of 0.2 ounces. Find the probability that a randomly selected box of cereal will weigh between 14.15 and 14.375 ounces. Round to the nearest tenth.(1 point)
Responses

19.7%
19.7%

97.0%
97.0%

77.3%
77.3%

95.0%

Answer 1

To find the probability that a randomly selected box of cereal will weigh between 14.15 and 14.375 ounces, we will use the properties of the normal distribution.

1. **Convert the weights to z-scores** using the formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where \( \mu \) is the mean and \( \sigma \) is the standard deviation.

Given:
- Mean \( \mu = 14 \) ounces
- Standard deviation \( \sigma = 0.2 \) ounces

**Calculate z-scores**:

For \( X = 14.15 \):
\[
z_1 = \frac{(14.15 - 14)}{0.2} = \frac{0.15}{0.2} = 0.75
\]

For \( X = 14.375 \):
\[
z_2 = \frac{(14.375 - 14)}{0.2} = \frac{0.375}{0.2} = 1.875
\]

2. **Look up the z-scores** in the standard normal distribution table or use a calculator:

- The probability corresponding to \( z_1 = 0.75 \):
\[ P(Z < 0.75) \approx 0.7734 \]

- The probability corresponding to \( z_2 = 1.875 \):
\[ P(Z < 1.875) \approx 0.9699 \]

3. **Calculate the probability of the range**:
\[
P(14.15 < X < 14.375) = P(Z < 1.875) - P(Z < 0.75)
\]
\[
P(14.15 < X < 14.375) \approx 0.9699 - 0.7734 = 0.1965
\]

4. **Convert to percentage**:
\[
0.1965 \times 100 \approx 19.7\%
\]

Therefore, the probability that a randomly selected box of cereal will weigh between 14.15 and 14.375 ounces is **19.7%**.

**Final Answer**: 19.7%