The weights of steers in a herd are distributed normally. The variance is 10,000 and the mean steer weight is 1400lbs. Find the probability that the weight of a randomly selected steer is between 1539 and 1580lbs. Round your answer to four decimal places.

And we can find this probability using the normal standard table with this difference:

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

Where and

We are interested on this probability

And we can solve the problem using the z score formula given by:

Using this formula we got:

And we can find this probability using the normal standard table with this difference:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?