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M-Studythis

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Can someone help me with this question:

[FONT=&quot]A standardized test that has been used by employees to screen job applicants is known to produce scores that are normally distributed with a mean of 70 and a standard deviation of 9. Suppose a particular firm requires applicants to score in the top 20% of test scores before being considered for employment. What score would a job applicant require to be considered by this firm? (Your answer should be correct to one decimal place.)[/FONT]
 

Drongoski

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The standardised variable has normal distribution N(0,1).

Read off normal distribution table: you find P(z>0.84) = 0.8 = 80%

So: solve for (x-70)/9 = 0.84 getting x = 77.56

So only applicants with scores exceeding 77.5 will be considered. (i.e. >= 77.6)



Edit

Why am I doing this student's uni assignment anyway?
 
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Timothy.Siu

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X~Normal(70, 9^2)
We want x, such that P(X>=x) = 0.2.
Now, u standardise X. P(Z >= (x-70)/9) = 0.2
1 - P(Z < (x-70)/9) = 0.2
P(Z<(x-70)/9) = 0.8
Now u just look in the Z table, for the value that satisfies that, and solve (x-70)/9 for it.
 

Drongoski

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Haha. Pierre - you are the expert. Give him the correct answer.
 

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