Question

The average time students need to finish a particular test is 72 minutes with a standard deviation of 12 minutes. (Assume that these times are normally distributed.)

1. What is the probability that a randomly selected student will finish the exam in more than 75 minutes?
2. If the professor wants 95% of students to have sufficient time to finish the test, how much time should she give them?
3. What is the probability that a group of 20 students will take on average more than 75 minutes to finish the exam?
4. What is the probability that in a group of 5 students, exactly 3 will finish the exam in more than 75 minutes?

Answer 1

a) P(X > 75)

= P((X - )/ > (75 - )/)

= P(Z > (75 - 72)/12)

= P(Z > 0.25)

= 1 - P(Z < 0.25)

= 1 - 0.5987

= 0.4013

b) P(X < x) = 0.95

or, P((X - )/ < (x - )/) = 0.95

or, P(Z < (x - 72)/12) = 0.95

or, (x - 72)/12 = 1.645

or, x = 1.645 * 12 + 72

or, x = 91.74

c) P( > 75)

= P(( - )/() > (75 - )/())

= P(Z > (75 - 72)/(12/))

= P(Z > 1.12)

= 1 - P(Z < 1.12)

= 1 - 0.8686

= 0.1314

d) n = 5

p = 0.4013

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X = 3) = 5C3 * (0.4013)^3 * (1 - 0.4013)^2

             = 0.2316