Question

The weights of steers in a herd are distributed normally. The standard deviation is

100lbs
100⁢lbs
and the mean steer weight is
1200lbs
1200⁢lbs
. Find the probability that the weight of a randomly selected steer is between
1000
1000
and
1369lbs
1369⁢lbs
. Round your answer to four decimal places.

Answer 1

0.9317

Explanation:

Standard deviation of the weights =
`\sigma`=100 lbs

Mean weight = u = 1200 lbs

We 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:

`z=(x-u)/(\sigma)`

1000 converted to z scores is:

`z=(1000-1200)/(100)=-2`

1369 converted to z scores is:

`z=(1369-1200)/(100)=1.69`

So, we have to find the values from z table that lie between -2 to 1.69

P( 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) = 0.9545

P(z < -2) = 0.0228

So,

P(-2 < z < 1.69) = 0.9545 - 0.0228 = 0.9317

Thus,

P( 1000 < x < 1369 ) = 0.9317

From 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