he time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions. (a) What is the probability of completing the exam in one hour or less? (Round your answer to four decimal places.) (b) What is the probability that a student will complete the exam in more than 60 minutes but less than 65 minutes? (Round your answer to four decimal places.) (c) Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time? (Round your answer up to the nearest integer.) students
a)
for normal distribution z score =(X-μ)/σ | |
here mean= μ= | 80 |
std deviation =σ= | 10.0000 |
probability = | P(X<60) | = | P(Z<-2)= | 0.0228 |
b)
probability = | P(60<X<65) | = | P(-2<Z<-1.5)= | 0.0668-0.0228= | 0.0440 |
c)
probability of taking more than 90 minutes:
probability = | P(X>90) | = | P(Z>1)= | 1-P(Z<1)= | 1-0.8413= | 0.1587 |
hence expected student that will be unable to complete the exam in the allotted time =np=60*0.1587 =9.522 ~10
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