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 115 seconds and a standard deviation of 20 seconds. The fastest 10% are to be given advanced training. What task times qualify individuals for such training

c = 89.4 seconds

The time for the fastest 10 % is less than 89.4 seconds

Step-by-step explanation:

The mean value is μ = 115 seconds and the standar deviation δ = 20 seconds.

We need to find c such that

P(x<c) = 0.1 It means, C is the time for the fastest 10 %

Such that:

We find z of the normal standard table

Thus

c = 89.4 seconds

The time for the fastest 10 % is less than 89.4 seconds

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316