Question

. what is the probatary that a single banery undendy selected bom the popdusen wil have Psis90) Round to four decimal places as needd b what is the probablity that 4 randonty sampled bateries from the P(70 s x % 90-[](Ro nd to four decir al places as needed ) c แ, between 70 and 90 hours? population wa have a sample mean 벼 of between 70 and 90 hours? the manufacture of the batlery is abls to reduce the standard deviation of baltery e fom 11 to 9 houts, what would be the probublity that 4 buteris randomly sanplied som the popution it hvs a sangle meam e betwoon 70 and 90 hours? Pssso)(Round to four decimal places as neoded)

Answer 1

Note that sample mean X follows a Normal distribution with Mean-μ-80 Standard Error-o/vn-11 Note that z-(x-μ)/(o/Vn) (a) -(x-80)/11 --(Eq.1) follows standard Normal. When-70, from (Eq.1,z -0.91 When-90, from (Eq.1),z 0.91 P(70X90)-P(-0.91Z0.91)- P(Z0.91)- P(Z-0.91)-0.8186-0.1814 0.6372 (ans) (b) Note that sample meanX follows a Normal distribution with Mean-μ-80 Standard Errors σ/vn :5.5 Note that z:(x-μ)/(o/vn) :(x-80)/55 When 70, from (Eq.1),z -1.82 --(Eq.1) follows standard Normal. When X-90, from (Eq1), z 1.82 P(7090)-P-1.82Z1.82)- P(Z 1.82)- P(Z-1.82)-0.9656-0.0344-0.9312 (an) (c) Note that sample mean X follows a Normal d Mean-μα 80 Standard Errors ơNn 4.5 Note that z- (-/o/Vn) x80)/4.5 (Eq1) follows standard Normal. When X 70, from (Eq.1), z-2.22 When 90, from (Eq.1), z 2.22 P(70 < X < 90)-P(-2.22 < Z <2.22-P(Z< 2.22)-P(Z <-2.22-0.9868-0.0132-09736 (ans)