Question

A livestock company reports that the mean weight of a group of young steers is 1115 pounds with a standard deviation of 80 pounds. Based on the model ​N(1115​,80​) for the weights of​ steers, what percent of steers weigh ​a) over 1100 ​pounds? ​b) under 1300 ​pounds? ​c) between 1150 and 1200 ​pounds?

Answer 1

A)

µ =    1115                  
σ =    80                  
                      
P ( X ≥   1100.00   ) = P( (X-µ)/σ ≥ (1100-1115) / 80)              
= P(Z ≥   -0.188   ) = P( Z <   0.188   ) =    0.5744   (answer)

B)

µ =    1115          
σ =    80          
              
P( X ≤    1300   ) = P( (X-µ)/σ ≤ (1300-1115) /80)      
=P(Z ≤   2.313   ) =   0.9896   (answer)

c)

µ =    1115                              
σ =    80                              
we need to calculate probability for ,                                  
P (   1150   < x <    1200   )                  
=P( (1150-1115)/80 < (X-µ)/σ < (1200-1115)/80 )                                  
                                  
P (    0.438   < Z <    1.063   )                   
= P ( Z <    1.063   ) - P ( Z <   0.438   ) =    0.8560   -    0.6691   =    0.1869 (ANSWER)

excel formula for probability from z score is =NORMSDIST(Z)