A certain type of automobile battery is known to last an average of 1140 days with a standard deviation of 80 days. If 400 of these batteries are selected, find the following probabilities for the average length of life of the selected batteries. (Round your answers to four decimal places.) (a) The average is between 1128 and 1140. (b) The average is greater than 1152. (c) The average is less than 940.

Answer:Mean = [tex]\mu = 1140[/tex]Standard deviation = [tex]\sigma = 80[/tex]Find the probabilities for the average length of life of the selected batteries.A)The average is between 1128 and 1140.We are supposed to fidn P(1128<x<1140)Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]At x = 1128 [tex]Z=\frac{1128-1140}{80}[/tex][tex]Z=-0.15[/tex]Refer the z table for p value P(x<1128)=0.4404At x = 1140 [tex]Z=\frac{1140-1140}{80}[/tex][tex]Z=0[/tex]Refer the z table for p value P(x<1140)=0.5So,P(1128<x<1140)=P(x<1140)-P(x<1128)=0.5-0.4404=0.0596Hence the probabilities for the average length of life of the selected batteries  is between 1128 and 1140 is 0.0596B)The average is greater than 1152.P(x>1152)At x = 1128 [tex]Z=\frac{1152-1140}{80}[/tex][tex]Z=0.15[/tex]Refer the z table for p value P(x<1152)=0.5596So,P(x>1152)=1-P(x<1152)=1-0.5596=0.4404Hence the probabilities for the average length of life of the selected batteries is greater than 1152 is 0.4404C) The average is less than 940.P(x<940)At x = 940 [tex]Z=\frac{940-1140}{80}[/tex][tex]Z=-2.5[/tex]Refer the z table for p value P(x<940)=0.5596Hence the probabilities for the average length of life of the selected batteries is less than 940 is 0.5596