Question

The variance for weight in a particular herd of cattle is 484 pounds2. The mean weight is 562 pounds. How heavy would an animal have to be if it was in the top 2.5% of the herd? The bottom 0.13%?

can someone explain how to do this? Thanks

Answer 1

Mean = 862 pounds

Variance = 484 pounds²

Standard deviation = = 22 pounds

Assuming the data is normally distributed, P(X < A) = P(Z < (A - mean)/standard deviation)

Let an animal be of weight T to be in top 2.5%

P(X > R) = 0.025

P(X < T) = 1 - 0.025

P(Z < (T - 862)/22) = 0.975

Take Z value corresponding to 0.025 from standard normal distribution table.

(T - 862)/22 = 1.96

T = 905 pounds   

Note: The ans would be 905 pounds if the empirical formula is used. i.e., z value is taken as 2

Let an animal be of weight B to be in bottom 0.13%

P(X < B) = 0.0013

P(Z < (B - 862)/22) = 0.0013

(B - 862)/22 = -3.01

B = 796 pounds