Question

In: Statistics and Probability

. IQ is normally distributed with a mean of 100 and a standard deviation of 15....

. IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose an individual is randomly chosen. a) (3pt) Find the probability that the person has an IQ greater than 125. b) (4pt) Find the probability that the person has an IQ score between 105 and 118. c) (4pt) What is the IQ score of a person whose percentile rank is at the 75th percentile, ?75? d) (3pt) Use the information from part (c) to fill in the blanks and circle the correct choice in the following statement. ________% of the individuals (persons) have IQ score less than/more than __________ e) (4pt) “MENSA” is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the “MENSA” organization

Solutions

Expert Solution

a)

µ =    100                          
σ =    15                          
right tailed                              
P ( X ≥   125.00   )                      
                              
Z =   (X - µ ) / σ = (   125.00   -   100   ) /    15   =   1.667
                              
P(X ≥   125   ) = P(Z ≥   1.667   ) =   P ( Z <   -1.667   ) =    0.0478

b)

µ =    100                              
σ =    15.000                              
we need to calculate probability for ,                                  
105   ≤ X ≤    118                          
X1 =    105   ,   X2 =   118                  
                                  
Z1 =   (X1 - µ ) / σ = (   105   -   100   ) /    15   =   0.3333  
Z2 =   (X2 - µ ) / σ = (   118   -   100   ) /    15   =   1.2000  
                                  
P (   105   < X <    118   ) =    P (    0.33333 < Z <    1.200   )
                                  
= P ( Z <    1.200   ) - P ( Z <   0.333   ) =    0.88493   -    0.630559   =    0.2544

c)µ =    100                  
σ =    15                  
proportion=   0.75                  
                      
Z value at    0.75   =   0.6745   (excel formula =NORMSINV(   0.75   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   0.674   *   15   +   100  
X   =   110.117 (answer)

d)

25% of the individuals (persons) have IQ score less than/more than 110.117

e)

µ =    100                  
σ =    15                  
proportion=   0.98                  
                      
Z value at    0.98   =   2.0537   (excel formula =NORMSINV(   0.98   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   2.054   *   15   +   100  
X   =   130.806(answer)   


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