Question

The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and standard deviation of 2 months. Find how long is an instrument that separates the bottom 83.398% will last? Round to the nearest hundredth.

Answer 1

To find the duration that separates the bottom 83.398% of the distribution, we need to determine the z-score that corresponds to the cumulative probability of 0.83398.

1. **Finding the z-score:**
- We can use a z-table, calculator, or statistical software to find the z-score corresponding to the cumulative probability of 0.83398.
- Looking it up, we find that the z-score for 0.83398 is approximately 0.967.

2. **Using the z-score formula:**
The z-score formula is given by:

\[
z = \frac{X - \mu}{\sigma}
\]

Where:
- \( z \) = z-score
- \( X \) = value in the distribution
- \( \mu \) = mean of the distribution (12 months)
- \( \sigma \) = standard deviation of the distribution (2 months)

3. **Solving for \( X \):**

Rearranging the formula to solve for \( X \):

\[
X = \mu + z \cdot \sigma
\]

Substituting in the values we have:

\[
X = 12 + 0.967 \times 2
\]

\[
X = 12 + 1.934
\]

\[
X \approx 13.934
\]

4. **Rounding:**
Rounding to the nearest hundredth, we have:

\[
X \approx 13.93
\]

Thus, the instrument that separates the bottom 83.398% of the distribution will last approximately **13.93 months**.