Question

A livestock company reports that the mean weight of a group of young steers is 1103 pounds with a standard deviation of 80 pounds. Based on the model ​N(1103​,80​) for the weights of​ steers, what percent of steers weigh

​a) over 1150 ​pounds?

​b) under 1300 ​pounds?

​c) between 1200 and 1250 ​pounds?

Answer 1

Solution :

Given that ,

mean = = 1103

standard deviation = = 80

a)

P(x > 1150) = 1 - P(x < 1150)

= 1 - P((x - ) / < (1150 - 1103) / 80)

= 1 - P(z < 0.5875)

= 1 - 0.7216   

= 0.2784 = 27.84%

percent = 27.84%

b)

P(x < 1300) = P((x - ) / < (1300 - 1103) / 80)

= P(z < 2.4625)

= 0.9931 = 99.31%

percent = 99.31%

c)

P(1200 < x < 1250) = P((1200 - 1103)/ 80) < (x - ) / < (1250 - 1103) / 80) )

= P(1.2125 < z < 1.8375)

= P(z < 1.8375) - P(z < 1.2125)

= 0.9669 - 0.8873

= 0.0796 = 7.96%

percent = 7.96%