Question

company xyz know that replacement times for the dvd players it produces are normally distributed with a mean of 5.8 years and a standard deviation of 2.4 years.
find the probability that a randomly selected dvd player will have a replacement time less than 0.8 years?
p(x < 0.8 years) =
enter your answer accurate to 4 decimal places. answers obtained using exact z - scores or z - scores rounded to 3 decimal places are accepted.
question help: post to forum

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 0.8\), \(\mu=5.8\), and \(\sigma = 2.4\).

$$z=\frac{0.8 - 5.8}{2.4}=\frac{-5}{2.4}\approx - 2.083$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt0.8)\), which is equivalent to \(P(Z\lt - 2.083)\) (since \(z=\frac{x-\mu}{\sigma}\)).
Looking up the value of \(P(Z\lt - 2.083)\) in the standard normal distribution table (or using a calculator with a normal - distribution function, e.g., in Excel: =NORM.S.DIST(-2.083,TRUE)).
\(P(Z\lt - 2.083)\approx0.0187\)

Answer:

\(0.0187\)