The mileage of a new model SUV on the interstate highway has been stated by the manufacturer as having a mean of 20 mpg and a standard deviation of 3.0 mpg. A consumer agency intends to sample 36 cars and test them. Find the probability that the sample average will be over 21.0 mpg. (Assume that mileage is normally distributed.)

Answer: 0.0227502Step-by-step explanation:Let x denote the random variable that represents the mileage of SUV.As per given we have,[tex]\mu = 20[/tex][tex]\sigma = 3[/tex]sample size : n= 36We assume that  the mileage of SUV is normally distributed.Z-score value corresponds to x= 21.0, [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}=\dfrac{21-20}{\dfrac{3}{\dfrac{}{36}}}=2[/tex]Using standard z-value table , The probability that the sample average will be over 21.0 mpg:-P(X>35)=P(z>2)=1-P(z<2)=1-0.9772498=0.0227502Hence, the required probability = 0.0227502