The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.9 cm a. Find the probability that an individual distance is greater than 217.50 cm.
b. Find the probability that the mean for 25 randomly selected distances is greater than 202.80 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
a. The probability is
(Round to four decimal places as needed.)

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.

can you post the graph?