Answer:
214.5 seconds or less time will qualify individuals for such training
Step-by-step explanation:
given that the time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean value of 150 sec and a standard deviation of 25 sec.
i.e. X the time that it takes a randomly selected job applicant to perform a certain task is N(150,25)
The fastest 10% are the ones who are above the 90th percentile
We know 90th percentile for Z std normal score is 1.28
Corresponding X score would be
[tex]x=\mu+1.28 \sigma\\X= 150+1.28(25)\\= 214.50[/tex]
Round off to one decimal
214.5 seconds or less time will qualify individuals for such training