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 what

range are 99.8% of the automobiles?

range=u ± 3.09 sd

Step-by-step explanation:

Given:

mean, u = 26.8 mpg

standard 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, - 195.68

range=u ± 3.09 sd = 249.28, - 195.68!