Question

The mean gas mileage for a hybrid car is 57 miles per gallon. Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 35 miles per on What proportion of lyrics pots ver miles per gallon? (b) What proportion of hybrids gets 53 miles per gallon or less (c) What proportion of hybrids gets between 57 and 61 miles per gallon? What is the probably that a randomly selected Hybrid gets less than 45 miles per gallon Click the icon to view a table of areas under the normal curve (a) The proportion of hybrids that gets over 61 miles per galon is (Round to four decimal places as needed) (b) The proportion of hybrids that gets 53 miles per gallon or less is (Round to four decimal places as needed) (c) The proportion of trybrids that gets between 57 and 61 miles per gallon is (Round to four decimal places as needed) (d) The probability that a randomly selected hybrid gets less than 45 miles per galon is. (Round to four decimal places as needed)

Answer 1

C = 3.5 Here we have to find that the p-value from the firen data The mean gas mullage for chybrid Standard deviation ► 2) P(X361) - P(Z > *) Plz > 6.4.-53) = P(2x. 4 Plz> > 1.143) 0.1263 >b) PCX 253) - P(Z2 53-57) - P/22 =P(Z <-1.143) = 10.1263 c) p (572x 261) 61-57 P(97-57_24 3.5 = PloL Z L PIOL 2 21.143) *P(2213) --P1240) = 0.8736-015 3.5 -) 3.5 y 3.5 :) 0.3736 3.5 » d) P(2245) = P(22 45-57 • P(22-12 -123.5) - Plz 2-3.45) - 10.0003

Sol we have to find that pralue, from the giren data mean u= 25cm Standard deviation r = 0.09 cm → a) P(x < 24.9) 24.9-25 0.09 PLZ 2-1,111) 0.1332 b) Pla4,61 2x 2 25.12) - Plz224,61 ) - P(Z 225.14) 5p/24,61-25 25.19.25 5.9-25) & Z 2 0.09 0.09 =P-4.4444)42 42.1111) :'P(22-4,4444) – PLZ 22.1111)

= P(Z 22-111) -PL22-4.4444) 0.9825985 c) we tercezi-0.9825985 0.0174 = 0.0174 5000 EX np 5000X 0.0174 get p ne Z 187 d) P= P(24.612x 295.19) :P(-Giuuuu 2 x 22.1111) 0.9825985 = np = 10000 X0.9825985 = 9825.985 tence 9825.985 Еz