Question

a manufacturer knows that their items have a normally distributed length, with a mean of 12.8 inches, and standard deviation of 2.4 inches.
if one item is chosen at random, what is the probability that it is less than 15.6 inches long?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 15.6\), \(\mu=12.8\), and \(\sigma = 2.4\).

$$z=\frac{15.6 - 12.8}{2.4}=\frac{2.8}{2.4}\approx1.17$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt15.6)\), which is equivalent to \(P(Z\lt1.17)\) (since \(X\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\) and \(Z=\frac{X - \mu}{\sigma}\)).
Looking up the value of \(z = 1.17\) in the standard - normal distribution table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: normalcdf(-100,1.17)), we find that \(P(Z\lt1.17)=0.8790\)

Answer:

\(0.8790\)