A company issues a general knowledge test to its potential employees as part of its recruitment procedures. The company does not invite the lowest scoring 20% back for a second interview. If the scores on this test are normally distributed with a mean of 60 and a standard deviation of 17. What mark (to the nearest whole number) must a candidate achieve in order to secure a second interview

Answer:

46

Step-by-step explanation:

Given :

Lowest scoring 20% ; this corresponds to score below (1 - 0.2) = 0.8

Mean, m = 60

Standard deviation, s = 17

Zscore = (x - mean) / standard deviation

Therefore, the Zscore for the bottom 20% equals ; P(x > Z) = 0.8

Using a Z probability calculator ;

The Zscore = -0.842

Therefore to Obtian the score, x

Zscore = (x - mean) / standard deviation

-0.842 = (x - 60) / 17

-0.842 * 17 = (x - 60)

-14.314 = x - 60

-14.314 + 60 = x

x = 45.686

To secure a second interview, candidate must score, 46 (nearest whole number).

The marks (to the nearest whole number) must a candidate achieve in order to secure a second interview is 46 and this can be determined by using the z-score formula.

Given :

Let 'x' be the mark must a candidate achieve in order to secure a second interview.

So, through the z-score formula, the value of 'x' can be determined.

[tex]\rm z_{score} = \dfrac{x-mean}{Standard\; Deviation}[/tex]

[tex]\rm -0.842 = \dfrac{x-60}{17}[/tex]

[tex]17\times -0.842 = x - 60[/tex]

x = 60 - 14.314

x = 45.686

So, the marks (to the nearest whole number) must a candidate achieve in order to secure a second interview is 46.

For more information, refer to the link given below:

https://brainly.com/question/17756962

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