Fuel economy estimates for automobilos built in a certain yoar predicted a mean of 26.8 mpg and a standard deviation of 72 mpg for highway driving. Assume that a normal distribution can be applied. Within whatrange are 99.8% of the automobiles?

Answer:range=u ± 3.09 sdStep-by-step explanation:Given:mean, u= 26.8 mpgstandard deviation, sd=72 mpg% contained in interval = 99.8%the interval for 99.8% of the values of a normal distribution is given by mean ± 3.09 standard deviation= u ± 3.09 sd                                                       =26.8  ± 3.09(72)                                                       =26.8 ± 222.48                                                       = 249.28 , -195.68range=u ± 3.09 sd = 249.28 , -195.68 !