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Assume the TV warranty or replacement times for TV sets are normally distributed with a mean of 9.2 years and a standard deviation of 1.1 years. Find the probability that a randomly selected TV will have a replacement time of less than 6 years.

Assume the TV warranty or replacement times for TV sets are normally distributed with-example-1

1 Answer

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Step-by-step explanation

Firstly, let's draw a picture of the distribution:

Graphically, we want to calculate the blue area. To put it differently, we want to calculate


P(X<6).

To do this, we need to find the z-score associated with 6. We can calculate it by


\begin{gathered} z=(6-\mu)/(\sigma)\Rightarrow\begin{cases}\mu=\operatorname{mean} \\ \sigma=\text{standard deviation}\end{cases}, \\ \\ z=(6-\mu)/(\sigma)=(6-9.2)/(1.1)\approx-2.91. \end{gathered}

Now, we must check our favorite z-score table and look for the probability associated with the z-score we just found. It's 0.0018.

Answer

The probability that a randomly selected TV will have a warranty of less than 6 years is 0.0018.

Assume the TV warranty or replacement times for TV sets are normally distributed with-example-1
User Schummbo
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