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:

$$\boxed{{\displaystyle 0.494}}$$