In: Statistics and Probability
During a fishing trip in the Rockies, you caught and released four red-spotted rocky mountain trout (Salvelinus malma) that had an average weight of 2.8 pounds. These trout are known to have an average weight of 3.6 pounds with a standard deviation of 1.7 pounds and we will assume that they are normally distributed. The percent of such samples of size four that have an average weight less than yours is closest to
SD for sample of 4 = SD (population) / (n^0.5)
= 1.7 / 2
= 0.85 pounds
X = mean of sample of 4 in pounds
we want P(X <= 2.8) :
percent of such samples of size four that have an average weight less than yours = 17.33%
(please UPVOTE)
P(z<Z) table :