The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 1300lbs. Find the probability that the weight of a randomly selected steer is between 939 and 1417lbs. Round your answer to four decimal places.

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()

Here, = population mean = 1300 lbs

          = population standard deviation = 200 lbs

The z score area distribution is given by;

            Z = ~ 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( < ) = P(Z < 0.58) = 0.71904

P(X <= 939) = P( <= ) = 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 .

ok

step-by-step explanation:

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98