The probability that the weight of a randomly selected steer is between 600 and 1060 lbs is 0.74364.
We are given that the weights of steers in a herd are distributed normally. The standard deviation is 200 lbs and the mean steer weight is 800 lbs.
Now,
Let X be the weights of steers in a herd,i.e.; X ~ N( μ, σ²)
Now,
Population mean, μ = 800 lbs
Population standard deviation, σ = 200 lbs
The z-score area distribution will be:
Z = [ ( X - μ )/σ ] ~ N(0,1)
Therefore, the probability that the weight of a randomly selected steer is between 600 and 1060 lbs
= P(600 lbs < X < 1060 lbs)
P(600 lbs < X < 1060 lbs) = P( X < 1060 ) - P( X ≤ 600 )
P(X < 1060) = P( ( X - μ )/σ < [ 1060 - 800 /200 ] ) = P( Z < 1.3) = 0.9023
P(X ≤ 600) = P( ( X - μ )/σ ≤ 600 - 800 /200 ) = P( Z ≤ - 1 ) = 1 - P( Z < 1 )
P(X ≤ 600) = = 1 - 0.84134 = 0.15866
Therefore, P( 600 lbs < X < 1060 lbs) = 0.9023 - 0.15866 = 0.74364
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