Answer:
range=u ± 3.09 sd
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 !