Question

a standard deviation of 15mpg Com In determining automobile mileage ratings, it was found that the mos O for a certain model is roma parts a through below Find PIX < 35)-2528 Round to four decimal places as needed) Find P133 << P(33<x<39- Round to four decimal places as needed) FPOX o und to four decimal places needed) did POS) P POX35) Find the wing that the per 15% of cars achieve The maighthea Round so we decimal places an d

Answer 1

Solution Given that mcan M 36 mpg Standard deviation :15 mpg P(X<35) =P(2-4 < 4536) =P(2575) =P(24-0.666667) = 0.2525 Thus, PCX635) = 6.2525 P(33 < X <39) **(*«Z<3) = P(-24242) = P(z<2) – P1242) = 0.9772-0.0228 -0.9544 Thus, P(33<x< 39) - 0.9544) PCX738)

P(2738) - P(2745) = P(271.3333) =1-P(2<1:3333) +/-0.9088 -0.0912 Thus, PCX738) = 0.09127 P(X735) *P(2775)

P(x>35) P(27-0.66667) +/- P(Z <-0166667) -1-0.2525 -0.7475 The mileage rating that the upper 156 cars achieve = x hence below x there will be 85% of data P-0185 2:1-036433 (from standaod normal table] we know that - x-u Z : - 9-036433 = 7 2-36 = 115546495 30 = 115546495+36 2 = 3755464950 hence, The mileage rating that the upper 157, of cars achieve is 37155 mpg