Suppose the following are the city driving gas mileages of a selection of sport utility vehicles (suvs). 19, 20, 19, 20, 18, 21, 17, 19, 24, 23, 21, 21, 17, 20, 20, 18 (a) find the sample standard deviation (rounded to two decimal places). (b) in what gas mileage range does chebyshev's inequality predict that at least 75% of the selection will fall? (enter your answer using interval notation. round your answers to two decimal places.) , (c) what is the actual percentage of suv models of the sample that fall in the range predicted in part (b)? (round your answer to two decimal places.) % which gives the more accurate prediction of this percentage: chebyshev's rule or the empirical rule? chebyshev's rule empirical rule

Step-by-step explanation:

The mean of the gas mileages is 317÷16=19.8125

317 is the sum total of all the figures and 16 is the number of figures in the distribution

Standard deviation is the square root of the variance and the variance is the mean of all squared deviations

The 16 squared deviations are

7.9102(×2) + 3.2852(×2) + 0.6602(×3) + 0.0352(×4) + 1.4102(×3) + 10.1602 + 17.5352 = 56.4382

56.4382÷16 = 3.5274

This is the Variance. The standard deviation is herefore √3.5274 =1.878 ~ 1.88 (to 2 decimal places)

(B) Chebyshev's inequality predicts that 75% of the selection will fall within 2 standard deviations of the mean

2×1.88=3.76

19.8125-3.76 = 16.05

19.8125+3.76= 23.57

The gas mileages are between 16.05 and 23.57

(C) the actual % of SUV models of the sample that fall in the above range is (15/16 × 100) = 93.75%

(D) the empirical rule gives the more accurate prediction

sorry, i don't know i just need points, i'm new here. chao