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 !

answer: i don't understand. what are the options

step-by-step explanation:

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