Question

Can you show how to draw the normal curve for each of the problems and label it as well?

Heights of MEN in the U.S. are normally distributed µ = 69.6 inches with σ = 3 inches.

-________ percent (to nearest %) of men in the U.S. are either shorter than 5 ft. or taller than 6 ft?

-In a group of 150 U.S. men, approximately ________ of them should be shorter than 65 inches.

-A male height of _______________ corresponds to the 58th percentile in the U.S. population. -_______________ is the cutoff height to be in the top 12% of male heights in the U.S.

-The middle 72% of U.S. men will be between ________ inches and ________ inches tall. -A man in the U.S. shorter than ___________ inches would be considered "unusually short. ( Can you Show your work or explain answer.)

Answer 1

µ =    69.6                              
σ =    3                              
P (   60.00   < X <   72.00   )                  
=P( (60-69.6)/3 < (X-µ)/σ < (72-69.6)/3 )                                  
                                  
P (    -3.200   < Z <    0.800   )                   
= P ( Z <    0.800   ) - P ( Z <   -3.20   ) =    0.7881   -    0.0007   =    0.7875

required probability = 1 - 0.7875 = 0.2125

b)

µ =    69.6      
σ =    3      
          
P( X < 65   ) = P( (X-µ)/σ ≤ (65-69.6) /3)  
=P(Z ≤   -1.53   ) =   0.0626

expected people = 150*0.0626 = 9.39 ≈ 10

c)

P(X≤x) =   0.5800                  
                      
Z value at    0.58   =   0.2019   (excel formula =NORMSINV(   0.58   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   0.202   *   3   +   69.6  
X   =   70.2057   (answer)          
A male height of ___70.21 in____________ corresponds to the 58th percentile in the U.S. population.

d)

P(X≤x) =   0.8800                  
                      
Z value at    0.88   =   1.1750   (excel formula =NORMSINV(   0.88   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.175   *   3   +   69.6  
X   =   73.1250   (answer)      

__73.13 in_ is the cutoff height to be in the top 12% of male heights in the U.S   

e)

proportion=   0.7200                      
proportion left    0.2800   is equally distributed both left and right side of normal curve                   
z value at   0.14   = ±   1.080   (excel formula =NORMSINV(   0.28   / 2 ) )  
                          
z = ( x - µ ) / σ                          
so, X = z σ + µ =                          
X1 =   -1.080   *   3   +   69.6   =   66.36
X2 =   1.080   *   3   +   69.6   =   72.84

f)

P(X≤x) =   0.0500                  
                      
Z value at    0.05   =   -1.6449   (excel formula =NORMSINV(   0.05   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.645   *   3   +   69.6  
X   =   64.67 (answer)      

A man in the U.S. shorter than ____64.67_______ inches would be considered "unusually short.
because probability of happening of this is less than 0.05