he following times were recorded by the quarter-mile and mile runners of a university track team (times are in minutes). Quarter-Mile Times: 0.91 0.96 1.05 0.94 1.02 Mile Times: 4.53 4.35 4.61 4.71 4.51 After viewing this sample of running times, one of the coaches commented that the quarter-milers turned in the more consistent times. Use the standard deviation and the coefficient of variation to summarize the variability in the data. Quarter-milers Standard deviation (to decimals) Coefficient of variation (to decimal) % Milers Standard deviation (to decimals) Coefficient of variation (to decimal) % Does the use of the coefficient of variation indicate that the coach's statement should be qualified

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step-by-step explanation:

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step-by-step explanation:

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i believe it's a.

i'm not sure on this one, most likely correct.

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.