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Window People Help x https:/Mwebcourses.niu.e x d Learn dav/pid-3831611-dt-content-rid-23001326 2/courses/20152-lsYE-335-----1MHw%234.pdf 3. The reaction time of a driver to visual stimulus is nor- mally distributed with a mean of 0.4 seconds and a standard deviation of 0.05 seconds. (a) What is the probability that a reaction requires more than 0.5 seconds? (b) What is the probability that a reaction requires between 0.4 and 0.5 seconds? (c) What is the reaction time that is exceeded 90% of the time?

Solution

Mean = u = 0.4
SD = 0.05

a) P(x > 0.5) :

x = 0.5

z = (x - u) / SD
z = (0.5 - 0.4) / 0.05
z = 2

We need P(z > 2)

USe this link ----> https://www.easycalculation.com/statistics/p-value-for-z-score.php

And since it asks for > 0.5, check the \"right tailed P-value\", which is 0.0228

Answer ---> 0.0228

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b) Between 0.4 and 0.5

z = (x - u) / SD
z = (0.4 - 0.4) / 0.05
z = 0

So, P(z < 0) = 0.5

z = (x - u) / SD
z = (0.5 - 0.4) / 0.05
z = 2

P(z < 2) = 0.9772

P(0 < z < 2) = 0.9772 - 0.5 = 0.4772

Answer ---> 0.4772

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When it is exceeded 90% of the time, it must be larger 10% of the time....

Look at this link ----> http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

And check the table for probability of 0.1

When P = 0.1, z = -1.28

z = (x - u) / SD

-1.28 = (x - 0.4) / 0.05

Crossmultiplying :

-1.28*0.05 = x - 0.4

x = 0.4 - 1.28*0.05

x = 0.336

So, a reaction time of 0.336 seconds is exceeded 90% of the time.... ---> ANSWER