Question

Monitors manufactured by TSI electronics have lifespans that have a normal distribution with a standard deviation of 1000 hours and a mean lifespan of 17,000 hours. If a monitor selected at random find the probability that the lifespan of the monitor would be more than 18,000 hours. Round answer to four decimal places

Answer 1

The given values are:

`\mu=17000,\sigma=1000,X=18000`

The z score is calculated as follows:

`\begin{gathered} z=(X-\mu)/(\sigma) \\ z=(18000-17000)/(1000) \\ z=1 \end{gathered}`

Thus the required probability is P(z>1)

Using the table it follows:

`P\left(x>Z\right)=0.1587`

Therefore the required probability is 0.1587