Ahiker recorded the temperature at different altitudes on a mountain trail. the results are shown in the table. altitude (feet) 1,500 2,800 3,500 4,500 5,000 6,000 8,000 8,500 9,000 9,500 temperature (°f) 75 73 72 70 66 63 60 60 57 55 calculate the approximate correlation coefficient for the data, and interpret the value to describe the correlation. weak positive linear correlation -0.986 strong negative linear correlation 0.025 strong postive linear correlation 0.986 weak negative linear correlation -0.025 interpretation of the correlation coefficient arrowright correlation coefficient of the data arrowright

b) the dogs need to shift their focus to human scent.

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.