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 the probability that an instrument produced by this machine will last less than 7 months.Round to the nearest thousandth.

Answer 1

To find the probability that an instrument produced by the machine will last less than 7 months, we will use the properties of the normal distribution. Given the mean (\(\mu\)) and standard deviation (\(\sigma\)):

- Mean (\(\mu\)) = 12 months
- Standard deviation (\(\sigma\)) = 2 months

We need to find the probability \(P(X < 7)\), where \(X\) is the normally distributed variable representing the length of life of the instrument.

First, we will standardize the variable \(X\) using the Z-score formula:

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

For \(X = 7\):

\[
Z = \frac{7 - 12}{2} = \frac{-5}{2} = -2.5
\]

Next, we can find the probability corresponding to this Z-score. We can look this up in the standard normal distribution table or use a calculator.

Using the Z-table, we find the probability that \(Z < -2.5\):

\[
P(Z < -2.5) \approx 0.0062
\]

Thus, the probability that an instrument produced by this machine will last less than 7 months is approximately:

\[
P(X < 7) \approx 0.0062
\]

Rounding this to the nearest thousandth, the final answer is:

\[
\boxed{0.006}
\]