My school has a 30-question mulitple choice test to place students in English courses. Each question has 4 choices (a,b,c,d). If you get 7 answers right, you pass. What is the statistical probability of getting 7 or more correct answers by filling out the answer sheet randomly? Do the chances of passing in this way make the test valid or invalid?
2007-02-06 00:22:52 · 6 answers · asked by holacarinados 4 in Science & Mathematics ➔ Mathematics
You can use Pascal's Triangle to determine the probability of randomly getting 0, 1, 2, ... 6 answers right, beginning p(0) = 0.25^30, p(1) = 30 * 0.25^29 * 0.75, p(2) = 435 * 0.25^28 * 0.75^2, . . . then add those up to get the random probability of failing with 6 or less correct answers, then subtract it from 1 to get the random probability of passing.
As an approximation, the distribution of random scores is binomial with a mean of 7.5 and a standard deviation of 2.37 (= sqrt (30 * 0.25 * 0.75)). About 68% of the corresponding normal distribution would be above the value 6.5, so that's my estimate of what the procedure in the first paragraph would give.
The test is clearly invalid.
2007-02-06 07:19:43 · answer #1 · answered by Anonymous · 0⤊ 0⤋
P of getting right the question: 1/4 = 25%
P of getting at least 7 questions right: 7/30 = 23%
P of passing the test randomly= (1/4) x (7/30) = 7/120= 6%
Validity of the test? 6 out of 100 will pass the test randomly in theory with some guts and good luck. I will say the test is good enough ust for placing students in an English course.
2007-02-06 01:04:26 · answer #2 · answered by Andres C 2 · 1⤊ 1⤋
The probabilityof getting all 50 questions ideal woild be (a million/4)^50 The risk of having 40 9 ideal could be (a million/4)^40 9 * (3/4) * 50, using fact that there are 50 possinilities for the incorrect one The risk of having 40 8 ideal could be (a million/4)^40 8 * ((3/4)^2) * 50 * 40 9/2, using fact that there are 50*40 9/2 procedures for the incorrect ones to be distruibuted in the 50 solutions The ... ... The risk of having 40 ideal could be (a million/4)^40 * ((3/4)^10) * (50! / (40!*10!), using fact that there are 50! / (39!*10!) procedures for the incorrect ones to be dispensed in the 50 solutions hence the risk of passing the try is the sum, from N=40 to N=50, of (a million/4)^N ((3/4)^(50-N)) * 50! / ( N! * (50-N)! ) which you're able to do in Excel difficulty-free sufficient, or via employing risk tables. the respond is very SMALL, of the comparable order as winning the 1st prize in the lottery.
2016-12-13 10:09:17 · answer #3 · answered by hergenroeder 4 · 0⤊ 0⤋
If you answered the questions completely at random you would get an average of 7.5 questions correct.
The laws of probability show that a monkey has more chance of passing than failing so it's not a particularly valid way of evaluating a child.
2007-02-06 00:36:58 · answer #4 · answered by Trevor 7 · 0⤊ 1⤋
P = probability of right answer = 1/4
Q = probability of wrong answer = ¾
n = 30
probability of passing
= sigma( 30CrP^r*Q^n-r), r = 7,8, .... 30
2007-02-06 00:40:39 · answer #5 · answered by Rhul s 2 · 0⤊ 0⤋
filling it out randomly? 25%, thats the chance you have on every question
2007-02-06 00:33:18 · answer #6 · answered by champers 5 · 0⤊ 2⤋