Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=111 and σ=18. (a) What proportion of children aged 13 to 15 years old have scores on this test above 101 ? (NOTE: Please enter your answer in decimal form. For example, 45.23% should be entered as 0.4523.) (b) Enter the score which marks the lowest 20 percent of the distribution. (c) Enter the score which marks the highest 15 percent of the distribution

Answer 1

a)

µ =    111                  
σ =    18                  
                      
P ( X ≥   101.00   ) = P( (X-µ)/σ ≥ (101-111) / 18)      

      
= P(Z ≥   -0.556   ) = P( Z <   0.556   ) =    0.7107   (answer)

b)

µ=   111                  
σ =    18                  
proportion=   0.2                  
                      
Z value at    0.2   =   -0.84 (excel formula =NORMSINV(   0.2   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.84   *   18   +   111  
X   =   95.85   (answer)          

c)

µ=   111                  
σ =    18                  
proportion= 1-.15= 0.85                  
                      
Z value at    0.85   =   1.04   (excel formula =NORMSINV(   0.85   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.04   *   18   +   111  

X   =   129.66   (answer)