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 between 7 and 12 months. Round to the nearest thousandth.

Answer 1

To find the probability that an instrument produced by the machine will last between 7 and 12 months, given that the life of the instrument follows a normal distribution with a mean (μ) of 12 months and a standard deviation (σ) of 2 months, we can use the Z-score formula:

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

where \(X\) is the value for which we want to find the Z-score.

### Step 1: Find the Z-scores

1. For \(X = 7\) months:
\[
Z_1 = \frac{(7 - 12)}{2} = \frac{-5}{2} = -2.5
\]

2. For \(X = 12\) months:
\[
Z_2 = \frac{(12 - 12)}{2} = \frac{0}{2} = 0
\]

### Step 2: Find the probabilities

Next, we will look up these Z-scores in the standard normal distribution table or use a calculator to find the probabilities.

1. For \(Z_1 = -2.5\):
- The probability \(P(Z < -2.5)\) gives us approximately \(0.0062\).

2. For \(Z_2 = 0\):
- The probability \(P(Z < 0)\) is \(0.5000\).

### Step 3: Calculate the probability between the Z-scores

To find the probability that the instrument lasts between 7 and 12 months, we find:

\[
P(7 < X < 12) = P(Z < 0) - P(Z < -2.5)
\]

Substituting the probabilities we found:

\[
P(7 < X < 12) = 0.5000 - 0.0062 = 0.4938
\]

### Final answer

Rounding to the nearest thousandth, the probability that an instrument produced by this machine will last between 7 and 12 months is:

\[
\boxed{0.494}
\]