Question

The time needed for students complete a certain paper and pencil maze follows an approximately Normal distribution and has a mean of 70 seconds and a standard deviation of 15 seconds.

a) If a student takes 100 seconds to complete the maze, what percentile would the student be?

b) What percentage of students complete the maze between 50 and 60 seconds?

Answer 1

Solution :

Given that ,

mean = = 70

standard deviation = = 15

P(X<100 ) = P[(X- ) / < (100 -70) /15 ]

= P(z <2 )

Using z table

= 0.9772

answer=97.72%

(B)

P(50< x < 60) = P[(50 -70) /15 < (x - ) / < (60 -70) / 15)]

= P( -1.33< Z < -0.67)

= P(Z <-0.67 ) - P(Z <-1.33 )

Using z table   

=0.2514-0.0918

=0.1596

answer=15.96%

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