Respuesta :
Answer:
Probability that the weight of a randomly selected steer is between 939 and 1417 lbs is 0.68389 .
Step-by-step explanation:
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 1300 lbs.
So, Let X = weights of steers in a herd ,i.e.; X ~ N([tex]\mu,\sigma^{2}[/tex])
Here, [tex]\mu[/tex] = population mean = 1300 lbs
[tex]\sigma[/tex] = population standard deviation = 200 lbs
The z score area distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
So, Probability that the weight of a randomly selected steer is between 939 and 1417 lbs = P(939 lbs < X < 1417 lbs)
P(939 lbs < X < 1417 lbs) = P(X < 1417) - P(X <= 939)
P(X < 1417) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{1417-1300}{200}[/tex] ) = P(Z < 0.58) = 0.71904
P(X <= 939) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{939-1300}{200}[/tex] ) = P(Z <= -1.81) = 1 - P(Z < 1.81)
= 1 - 0.96485 = 0.03515
Therefore, P(939 lbs < X < 1417 lbs) = 0.71904 - 0.03515 = 0.68389 .
