Question

A livestock company reports that the mean weight of a group of young steers is 1142 pounds with a standard deviation of 81 pounds. Based on the model ​N(1142,81) for the weights of​ steers, what percent of steers weigh . Round answers to one decimal place.

a) over 1200 ​pounds? ​

b) under 1100 ​pounds?

​c) between 1000and 1150 ​pounds?

Answer 1

Solution :

Given that,

mean = = 1142

standard deviation = = 81

P(x >1200 ) = 1 - P(x<1200 )

= 1 - P[(x -) / < (1200-1142) / 81]

= 1 - P(z <0.72 )

Using z table

= 1 -  0.7642

=0.2358

=23.58%

B

P(x<1100 )

= P[(x -) / < (1100-1142) / 81]

= P(z <-0.52)

Using z table

=0.3015

=30.15%

C

P(1000< x < 1150) = P[(1000-1142) / 81 < (x - ) / < (1200-1150) / 81)]

= P( -1.75< Z < 0.62)

= P(Z < 0.62) - P(Z <-1.75 )

Using z table   

=  0.7324 - 0.0401

= 0.6923

=69.23%