Question

Exam scores for a population were standardized with a population mean (µ) of 500 and a population standard deviation (σ) of 100 on both the Math and Verbal portions. The following questions refer to the Math portion;
a) A highly selective school decides that it will consider only applicants with a score in the top 5%. What will be the minimum score that you must have in order to be considered for admission to this school?
b) If 1 million students took the exam, how many of these students would be viable applicants to this school?
4a. The mean weight of 140 6th graders is 80 lbs with a standard deviation of 8 lbs. Calculate the standard error of the mean.
b. Using the mean and standard deviation from Q4a, calculate the proportion (percentage) of 7th graders that will have weights between 74 lbs and 84 lbs?

Answer 1

a)

µ =    500                  
σ =    100                  
top 5% = bottom 95% or   0.9500                  
                      
Z value at    0.95   =   1.6449   (excel formula =NORMSINV(   0.95   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.645   *   100   +   500  
X   =   664.49             
minimum score = 664.49

b)

total students would be viable applicants to this school = 5% of 1 million = 50000 students

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4)

a)

µ =    80
σ =    8
n=   140

std error of mean =σ/√n = 8/√140 = 0.6761

b)

we need to calculate probability for ,                                          
74   ≤ X ≤    84                                  
X1 =    74   ,    X2 =   84                          
                                          
Z1 =   (X1 - µ )/(σ/√n) = (   74   -   80   ) / 0.6761=   -8.87  
Z2 =   (X2 - µ )/(σ/√n) = (   84   -   80   ) / 0.6761 =   5.92  
                                          
P (   74   < X <    84   ) =    P (    -8.87   < Z <    5.92   )       
                                          
= P ( Z <    5.92   ) - P ( Z <   -8.87   ) =    1.0000   -    0.0000   =    1.0000       (answer)
excel formula for probability from z score is =NORMSDIST(Z)