Answer:
The task time for the fastest 10% should be greater than 161.9 seconds.
Step-by-step explanation:
Since it is given that:
Mean = 130 seconds
Standard Deviation = 25 seconds
The fastest of the 10% can be found by finding the area corresponding to the required normal distribution curve
Thus the Z value for the normal distribution curve for the given values of mean and deviation is
Z = 1.2796
Thus the value of score can be found from the equation
[tex]Z=\frac{X-\overline{X}}{\sigma }[/tex]
Applying values we get
[tex]1.2796=\frac{X-130}{25}\\\\\therefore X=161.99[/tex]