A large company insists that all job applicants who are invited for interview ta

Question

A large company insists that all job applicants who are invited for interview take a psychometric test. The results of these tests follow a normal distribution with a mean of 61% and the standard deviation of 7.2%.

(a) What proportion of applicants would be expected to score over 70%?

(b) What proportion of applicants would be expected to score under 40%?

(c) What proportion of applicants would be expected to score between 50% and 65%?

(d) What score is exceeded by 20% of applicants?

(e) What is the highest score achieved by the 5% of applicants who do least well in the test?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    70      
u = mean =    61      
          
s = standard deviation =    7.2      
          
Thus,          
          
z = (x - u) / s =    1.25      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.25   ) =    0.105649774 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    40      
u = mean =    61      
          
s = standard deviation =    7.2      
          
Thus,          
          
z = (x - u) / s =    -2.916666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.916666667   ) =    0.001768968 [ANSWER]

*******************

c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    50      
x2 = upper bound =    65      
u = mean =    61      
          
s = standard deviation =    7.2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.527777778      
z2 = upper z score = (x2 - u) / s =    0.555555556      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.063283861      
P(z < z2) =    0.710742639      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.647458779   [ANSWER]

******************

d)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.80 =   0.2      
          
Then, using table or technology,          
          
z =    -0.841621234      
          
As x = u + z * s,          
          
where          
          
u = mean =    61      
z = the critical z score =    -0.841621234      
s = standard deviation =    7.2      
          
Then          
          
x = critical value =    54.94032712   [ANSWER]

*********************

e)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.05      
          
Then, using table or technology,          
          
z =    -1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    61      
z = the critical z score =    -1.644853627      
s = standard deviation =    7.2      
          
Then          
          
x = critical value =    49.15705389   [ANSWER]  
  
     

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