Question

IQ Scores are normally distributed with a mean of 100 and a standard deviation of 15. Use this information and a Z-table (or calculator or Excel) to solve the following problems.

A. Ryan has an IQ score of 118. Calculate the z-score for Ryan's IQ.

B. Interpret the z-score you obtained in the previous problem.

C. Suppose an individual is selected at random. What is the probability that their IQ score is less than 111? Round your answer to four decimal places (since this is what is given in the z-table)

D. Suppose an individual is selected at random. What is the probability that their IQ score is greater than 122? Round your answer to four decimal places.

E. Suppose an individual is selected at random. What is the probability that their IQ score is between 95 and 112? Round your answer to four decimal places.

F. MENSA is an organization you can join if you have a high IQ score. MENSA only admits people with IQ scores in the top 2% of people. What is the minimum IQ score you can get and still be admitted to MENSA?  Round your answer to the nearest whole number.

Answer 1

a)

µ=   100
σ=   15
X=   118
Z=(X-µ)/σ=   (118-100)/15)=       1.2

b)

A positive z-score indicates the raw score is higher than the mean average.

c)

µ =    100      
σ =    15      
          
P( X ≤    111   ) = P( (X-µ)/σ ≤ (111-100) /15)  
=P(Z ≤   0.733   ) =   0.7683
          
excel formula for probability from z score is =NORMSDIST(Z)          

...........

d)

µ =    100                  
σ =    15                  
                      
P ( X ≥   122.00   ) = P( (X-µ)/σ ≥ (122-100) / 15)              
= P(Z ≥   1.467   ) = P( Z <   -1.467   ) =    0.0712   (answer)

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

...........

e)

µ =    100                                  
σ =    15                                  
we need to calculate probability for ,                                      
P (   95   < X <   112   )                      
=P( (95-100)/15 < (X-µ)/σ < (112-100)/15 )                                      
                                      
P (    -0.333   < Z <    0.800   )                       
= P ( Z <    0.800   ) - P ( Z <   -0.333   ) =    0.7881   -    0.3694   =    0.4187   (answer)
excel formula for probability from z score is =NORMSDIST(Z)   

.........

f)

µ=   100                  
σ =    15                  
proportion=   0.98                  
                      
Z value at    0.98   =   2.05   (excel formula =NORMSINV(   0.98   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   2.05   *   15   +   100  
X   =   130.81  

= 131  (answer)          

minimum IQ score you can get and still be admitted to MENSA = 131

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

Please revert back in case of any doubt.

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