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SECTION 9.2 7. The Environmental Protection Agency (EPA) fuel economy estimates for automobiles suggest a mean of 24.8 mpg and a standard deviation of 6.2 mpg for highway driving Assume that a Normal model can be applied 132 136 137 a) Draw the model for auto fuel economy Clearly label it, showing what the 68-95-99.7 Rule predicts about miles per gallon b) in what interval would you expect the central 68% of autos to be found? About what percent of autos should get more than 31 mpe? d) About what percent of cars should get between 31 and 37 2mpe? e) Describe the gas mileage of the worst 25% of all cars The binomial model has and obelow: mean, 24.8 mpg 148 Standard Deviation, 6.20 mpe 150 6.20 1 2.40 18.60 3100 3720 43.40 24.80 68% 95% 99.7% 15) b) we would expect the central 68% of automobiles To be within the interval of 18.6 to 31.0 mpg. 159 c) About what percent of autos should get more than 31 mpg? 160 PIZ > 1) - P(x > 31 mpe), about 16% of automobiles should get more than 31 mpe Pl-1<Z<+1) = P(18.60<x<31.00) = 68% the distribution is symmetric, then 1/2 of (1-68%) is above 31 mpe PIX>31) = 16% PZI) -0.8413 PIZ > +1) - PDX > 31 mpe) - 15.87% 112 d) About what percent of cars should get between 31 and 37.2 mpg? P[+1 <Zc+2) = P131.00<x<37.20) - 13.59% PIZ>2) - PIZ> 1) = P(+1 <Z<+2) = 13.59% 175 177 178 179 e) Describe the gas mileage of the worst 2.5% of all cars. PIZ-2) = PDXC 12.40 mpg) = 2.28% 180 181

the 1 may not be correct

Answer 1

b, within the intervals between 18.6 mpg and 31 mpg

af The binomial model has u and & below: Meanu, = 24.8 mpg 2 =Cy-y) Standard deviation & = 6-20 meg z 1-3 -2 H 0 1 2 3 x 6.20 12.40 18.60 24.80 31 37.20 43 20 68%. 95% 99.9% T-68.1.1 - 91544 --- 99.74 - 6.2 12-24 18.6 2408 31 37.2 43.29

C,
C; 2= Y% - 31-24.8 62 nocna car = 0.1587 dy z = 2 % = 3702-24 820 6.2 noomal ede = 0.1359
e,
P (z <^2) = P(x412.4 mag) = 2.28%. To be at 12-4 meg and smaller PCZ-2 sampg and