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?

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  • M3
    Lv 7
    9 years ago
    Favorite Answer

    P(x < 17) = P(z < (17-18.7)/2.1) = P(z < -0.81) = 20.9% <------

  • 9 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.

  • Anonymous
    9 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

    About 21%.

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