Question

1. (4 points) the length of life of an instrument produced by a machine has a normal distribution with mean life of 12 months, and a standard deviation of 2 months. find the probability that an instrument produced by the machine will last between 9 and 12 months.

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 12\) (mean), \(\sigma=2\) (standard deviation).
For \(x = 9\):
\(z_1=\frac{9 - 12}{2}=\frac{-3}{2}=-1.5\)
For \(x = 12\):
\(z_2=\frac{12 - 12}{2}=0\)

Step2: Use the standard normal distribution table

We want to find \(P(9<X<12)\), which is equivalent to \(P(-1.5<Z<0)\) in the standard normal distribution (\(Z\)).
We know that \(P(-1.5<Z<0)=P(Z < 0)-P(Z<-1.5)\)
From the standard normal distribution table, \(P(Z < 0)=0.5\) and \(P(Z<-1.5)=0.0668\)

Step3: Calculate the probability

\(P(-1.5<Z<0)=0.5 - 0.0668=0.4332\)

Answer:

The probability that an instrument produced by the machine will last between 9 and 12 months is \(0.4332\)