Question

Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 5.9-in and a standard deviation of 0.8-in.

In what range would you expect to find the middle 50% of most head breadths?

Between and .

If you were to draw samples of size 41 from this population, in what range would you expect to find the middle 50% of most averages for the breadths of male heads in the sample?

Between and .

Enter your answers as numbers. Your answers should be accurate to 2 decimal places.

Answer 1

µ =    5.9                      
σ =    0.8                      
proportion=   0.5000                      
proportion left    0.5000   is equally distributed both left and right side of normal curve                   
z value at   0.25   = ±   0.674   (excel formula =NORMSINV(   0.50   / 2 ) )  
                          
z = ( x - µ ) / σ                          
so, X = z σ + µ =                          
X1 =   -0.674   *   0.8   +   5.9   =   5.36
X2 =   0.674   *   0.8   +   5.9   =   6.44
...................

µ =    5.9                              
σ =    0.8                              
n=   41                              
proportion=   0.50                              
proportion left    0.50   is equally distributed both left and right side of normal curve                           
z value at   0.25   = ±   -0.67   (excel formula =NORMSINV(   0.50   / 2 ) )          
Z value at    0.75   =   0.67                      
z = ( x - µ ) / (σ/√n)                                  
so, X = z σ / √n + µ =                                  
X1 =   -0.67   *   0.8   / √   41   +   5.9   =   5.82
X2 =   0.67   *   0.8   / √   41   +   5.9   =   5.98

...................................

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