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:
\[ \boxed{0.006} \]
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