Question

A livestock company reports that the mean weight of a group of young steers is 1115 pounds with a standard deviation of 68 pounds. Based on the model ​N(1115​,68​) for the weights of​ steers, what percent of steers weigh ?

​a) over 1050 ​pounds? ​

b) under 1000 ​pounds? ​

c) between 1250 and 1300 ​pounds?

Answer 1

Solution :

Given that ,

mean = = 1115

standard deviation = = 68

a) P(x > 1050) = 1 - p( x< 1050)

=1- p P[(x - ) / < (1050 - 1115) / 68]

=1- P(z < -0.96)

Using z table,

= 1 - 0.1685

= 0.8315

The percentage = 83.15%

b) P(x < 1000) = P[(x - ) / < (1000 - 1115) / 68 ]

= P(z < -1.69)

Using z table,

= 0.0455

The percentage = 4.55%

c) P(1250 < x < 1300) = P[(1250 - 1115)/ 68) < (x - ) /  < (1300 - 1115) /68 ) ]

= P(1.99 < z < 2.72)

= P(z < 2.72) - P(z < 1.99)

Using z table,

= 0.9967 - 0.9767

= 0.0200

The percentage = 2%