The quality control manager at a computer manufacturing company believes that the mean life of a computer is 83 months, with a standard deviation of 9 months. If he is correct, what is the probability that the mean of a sample of 78 computers would be less than 82.06 months? Round your answer to four decimal places.

Answer: 0.1788Step-by-step explanation:As per given , we have[tex]\mu=83,\ \ \sigma=9[/tex]Let x be the random variable that represents the life of a computer.Sample size : n= 78 Then, the z-score corresponds to x= 82.06 will be :-[tex]z=\dfrac{82.06-83}{\dfrac{9}{\sqrt{78}}}[/tex] [∵ [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]][tex]z=-0.92242835715\approx-0.92[/tex]Required probability ( using standard z-value table ) :-[tex]P(x<82.06)=P(z<-0.92)=1-P(z<0.92)\\\\=1-0.8212136=0.1787864\approx0.1788[/tex]Hence, the probability that the mean of a sample of 78 computers would be less than 82.06 months = 0.1788