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: 0.9032Step-by-step explanation:Given: Mean : [tex]\mu = 20,000\text{ hours}[/tex]Standard deviation : [tex]\sigma = 1800 \text{ hours}[/tex]The formula to calculate z is given by :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x= 17659 [tex]z=\dfrac{17659-20000}{1800}=−1.30055555556\approx-1.3[/tex]The P Value =[tex]P(z>-1.5)=1-P(z<1.3)=1- 0.0968005\approx0.9031995\approx 0.9032[/tex]Hence, the probability that the life span of the monitor will be more than 17,659 hours = 0.9032