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.

1 answer

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.