The weights of steers in a herd are distributed normally. The standard deviation is 100lbs100⁢lbs and the mean steer weight is 1200lbs1200⁢lbs. Find the probability that the weight of a randomly selected steer is between 10001000 and 1369lbs1369⁢lbs. Round your answer to four decimal places.

Answer:0.9317Step-by-step explanation:Standard deviation of the weights = [tex]\sigma[/tex]=100 lbsMean weight = u = 1200 lbsWe need to find the probability that the weight(x) of a randomly selected steer is between 1000 lbs and 1369 lbs i.e. P(1000 < x < 1369)Since, weights follow the normal distribution we can use the z values to find the required weight. For this we have to convert both the values to z score. The formula for z scores is:[tex]z=\frac{x-u}{\sigma}[/tex]1000 converted to z scores is:[tex]z=\frac{1000-1200}{100}=-2[/tex]1369 converted to z scores is:[tex]z=\frac{1369-1200}{100}=1.69[/tex]So, we have to find the values from z table that lie between -2 to 1.69P( 1000 < x < 1369 ) = P(-2 < z < 1.69)P(-2 < z < 1.69) = P(z < 1.69) - P(z < -2) From the z table:P(z < 1.69) = 0.9545P(z < -2) = 0.0228So,P(-2 < z < 1.69) = 0.9545 - 0.0228 = 0.9317Thus,P( 1000 < x < 1369 ) = 0.9317From this we can conclude that:The probability that the weight of a randomly selected steer is between 1000 lbs and 1369 lbs is 0.9317