Question

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a standard deviation of 1800 hours and a mean life span of 20,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 17,659 hours. Round your answer to four decimal places.

Answer 1

Answer: 0.9032

Step-by-step explanation:

Given: Mean : $\mu = 20,000\text{ hours}$

Standard deviation : $\sigma = 1800 \text{ hours}$

The formula to calculate z is given by :-

$z=\dfrac{x-\mu}{\sigma}$

For x= 17659

$z=\dfrac{17659-20000}{1800}=−1.30055555556\approx-1.3$

The P Value =$P(z>-1.5)=1-P(z<1.3)=1- 0.0968005\approx0.9031995\approx 0.9032$

Hence, the probability that the life span of the monitor will be more than 17,659 hours = 0.9032