The length of time taken by students on a statistics exam is normally distributed with a mean of 45 minutes and a standard deviation of 3 minutes. What proportion of students will finish the exam in 50 minutes or less?
Solution :
Given that,
mean =
= 45
standard deviation =
=3
P(X<50 ) = P[(X-
) /
< (50-45) /3 ]
= P(z < 1.67)
Using z table
=0.9525
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