probability questions

Question

You have been provided with marks obtained by 1000 students of mass communication. the average score of student was 68 with a standard deviation of 16. assuming the data to be normally distributed score of a student?

a: is more than 45

b: between 68 and 84

c: is less than 68

d: between 41 and 75

e: is less than 28

Answer 1

Note: Because sample size is 1000 we can use the normal distribution instead of the student t distribution with the assumption that mu = 68 and sigma = 16

Two methods. One Z score based (standard normal distribution), One Normal Cumulative Distribution function based.

Z score based

a)Determine Probability P that a randomly selected student will have a score x greater than 45 [ P(x>45) ] Z = (45-68)/16 = -1.4378 P(Z>-1.4378) = .9247

b) P(68<x<84) [since mu=68, this is the same as P(x<84) - .5 as P(x<68) = .5

Z = (84-68) /16 = 1 P(Z<1) = .8413 answer = .8413 - .5 = .3413

c) P(x<68) - answer was given in b = .5

d) P(41<x<75) = P(x<75) - P(x<41) Z(41) = (41-68)/16 = -1.688 Z(75) = (75-68)/16 = .4375

P(-1.688 <Z < .4375) = .6691 - .0457 = .6234

e) P(x<28) Z = (28-68)/16 = -2.50 P(Z< -2.50) = .0062

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Second method (using norm cumulative distribution function on TI-8x series calculator

a) P(x>45) = nmcdf(45,9e6,68,16) = .9247

b) P(68<x<84) = nmcdf(68,84,68,16) = .3413

c) P(x<68) = nmcdf(-9e6,68,68,16) = .5

d) P(41<x<75) = nmcdf(41,75,68,16) = .6234

e) P(x<28) = nmcdf(-9e6,28,68,16) = .0062

Moral, if you have a scientific calculator with normal cumulative distribution function available, use it.