Question

The weights of steers in a herd are distributed normally. The variance is 90,000 and the mean steer weight is 1500 lbs. Find the probability that the weight of a randomly selected steer is between 1859 and 2009 lbs. Round your answer to four decimal places.

Answer 1

solution

Given that ,

mean = = 1500

standard deviation = = 90000=300

P(1859< x <2009 ) = P[(1859-1500) /300 < (x - ) / < (2009-1500) / 300)]

= P( 1.20< Z < 1.70)

= P(Z <1.70 ) - P(Z <1.20 )

Using z table   

=  0.9554-0.8849

probability= 0.0705