Question

Multiple question needing guidance please.

1. A distribution of values is normal with a mean of 173.2 and a standard deviation of 39.
Find the probability that a randomly selected value is between 200.5 and 298.
P(200.5 < X < 298) =_______________

2. A distribution of values is normal with a mean of 229.7 and a standard deviation of 83.5.
Find P₃₂, which is the score separating the bottom 32% from the top 68%.
P₃₂ = __________________

3. 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 1-in. Due to financial constraints, the helmets will be designed to fit all men except those with head breadths that are in the smallest 2.4% or largest 2.4%.
What is the minimum head breadth that will fit the clientele?
min = ________________
What is the maximum head breadth that will fit the clientele?
max = __________________
Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

1)

µ =    173.2                                  
σ =    39                                  
we need to calculate probability for ,                                      
P (   200.5   < X <   298   )                      
=P( (200.5-173.2)/39 < (X-µ)/σ < (298-173.2)/39 )                                      
                                      
P (    0.700   < Z <    3.200   )                       
= P ( Z <    3.200   ) - P ( Z <   0.700   ) =    0.9993   -    0.7580   =    0.2413   (answer)

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

2)

µ=   229.7                  
σ =    83.5                  
proportion=   0.32                  
                      
Z value at    0.32   =   -0.468 (excel formula =NORMSINV(   0.32   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.468 *   83.5   +   229.7  
X   =   190.6 (answer)          

3)

µ=   5.9                  
σ =    1                  
proportion=   0.024                  
                      
Z value at    0.024   =   -1.977 (excel formula =NORMSINV(   0.024   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.977 *   1   +   5.9  
X   =   3.923   (answer)      

minimum head breadth that will fit the clientele = 3.9 in  

---------------------------------

µ=   5.9                  
σ =    1                  
proportion= 1-0.024 = 0.976                  
                      
Z value at    0.976   =   1.977 (excel formula =NORMSINV(   0.976   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.977 *   1   +   5.9  
X   =   7.877   (answer)          
maximum head breadth that will fit the clientele = 7.9 in