Question

For an adult male the average weight is 179 pounds and the standard deviation is 29.4 pounds. We can assume that the distribution is Normal (Gaussian). Answer the following questions either via simulations (use 10000 points) or via “rule of thumbs”. I). What is the approximate probability that a randomly picked adult male will weigh more than 180 pounds? Pick the closest answer. (6.66 points) a. About 15% b. About 30% c. About 50% d. About 65% II) What would be the range [A to B], which would contain about 68% of the adult males? Pick the closest answer. (6.66 points) a. Between 149.6 and 208.4 pounds b. Between 155.5 and 211.4 pounds c. Between 120.2 and 237.8 pounds d. Between 130.1 and 233.2 pounds III) A random adult male is chosen, which of these outcomes is the least likely? (6.66 points) a. The person will weigh more than 210 pounds b. The person will weigh more than 225 pounds c. The person will weigh less than 150 d. The person will weigh between 170 and 200 pounds

Answer 1

a)

µ =    179                          
σ =    29.4                          
right tailed                              
P ( X ≥   180.00   )                      
                              
Z =   (X - µ ) / σ = (   180.00   -   179   ) /    29.4   =   0.034
                              
P(X ≥   180   ) = P(Z ≥   0.034   ) =   P ( Z <   -0.034   ) =    0.4864

so, answer is c. About 50%

b)

µ =    179                          
σ =    29.400                          
proportion=   0.68                          
proportion left    0.32   is equally distributed both left and right side of normal curve                       
z value at   0.16   = ±   -0.9945   (excel formula =NORMSINV(   0.32   / 2 ) )      
Z value at    0.84   =   0.9945                  
z = ( x - µ ) / σ                              
so, X = z σ + µ =                              
X1 =   -0.99   *   29.40   +   179   =   149.76  
X2 =   0.99   *   29.40   +   179   =   208.24  

so, answer is Between 149.6 and 208.4 pounds

c)

µ =    179                          
σ =    29.4                          
right tailed                              
P ( X ≥   225.00   )                      
                              
Z =   (X - µ ) / σ = (   225.00   -   179   ) /    29.4   =   1.565
                              
P(X ≥   225   ) = P(Z ≥   1.565   ) =   P ( Z <   -1.565   ) =    0.0588

so, answer is

b. The person will weigh more than 225 pounds is least likely

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