Help with normal distribution?
The number of bristles on the fifth abdominal sternite in male flies was shown to follow a normal distribution with mean 18.7 and standard dev 2.1.
What percentage of the male population has fewer than 17 abdominal bristles?
- M3Lv 79 years agoFavorite Answer
P(x < 17) = P(z < (17-18.7)/2.1) = P(z < -0.81) = 20.9% <------
- Captain BrockLv 49 years ago
That is crazy. But it looks like those with 17 bristles lie at (18.7 - 17) / 2.1, or 0.8095 standard deviations away from the mean. Using a Standard Normal Distribution Table, we see that 29% have 17 or more (on one side of the distribution curve.) Subtract this from 50% (the lower one-half of the curve) and you get 21%.
Your poor little avatar looks so bewildered, I hope you can get through this and do something else that you enjoy.
- Anonymous9 years ago
Let X = # of abdominal bristles.
P( X<17 ) = P( (X-18.7)/2.1 < ( 17-18.7)/2.1 ) = P( Z < -0.8095238 ) = 0.2091070