Complete question is;
Researchers from the Educational Testing Service (ETS) found that providing immediate feedback to students answering openended questions can dramatically improve students’ future performance on exams (Educational and Psychological Measurement, Feb. 2010). The ETS researchers used questions from the Graduate Record Examination (GRE) in the experiment. After obtaining feedback, students could revise their answers. Consider one of these questions. Initially, 50% of the students answered the question correctly. After providing immediate feedback to students who answered incorrectly, 70% answered correctly. Consider a bank of 100 open-ended questions similar to those on the GRE. a. In a random sample of 20 students, what is the probability that more than half initially answer the question correctly? b. Refer to part a . After providing immediate feedback, what is the probability that more than half of the students answer the question correctly?
Answer:
A) 41.2%
B) 95.2%
Explanation:
This is a binomial probability distribution problem, so we will use the formula;
P(k) = (n!/(k!(n - k)!) × p^(k) × (1 - p)^(n-k)
A) Initially, 50% of the students answered the question correctly.
Thus, p = 0.5
Also,n = 20
Now, the probability that more than half initially answer the question correctly would be:
P(k > 10)
This can be expressed as;
P(k > 10) = P(11) + P(12) + P(13) + P(14) + P(15) + P(16) + P(17) + P(18) + P(19) + P(20)
P(11) = (20!/(11!(20 - 11)!) × 0.5^(11) × (1 - 0.5)^(20-11) = 0.1602
Similarly,
P(12) = (20!/(12!(20 - 12)!) × 0.5^(12) × (1 - 0.5)^(20-12) = 0.1201
P(13) = (20!/(13!(20 - 13)!) × 0.5^(13) × (1 - 0.5)^(20-13) = 0.0739
Using online binomial probability calculator we can get the remaining values which are;
P(14) = 0.037
P(15) = 0.0148
P(16) = 0.0046
P(17) = 0.0011
P(18) = 0.0002
P(19) = 0.00002
P(20) = 0
Thus;
P(k > 10) = 0.1602 + 0.1201 + 0.0739 + 0.037 + 0.0148 + 0.0046 + 0.0011 + 0.0002 + 0.00002 + 0 ≈ 0.41192 = 41.2%
B) After providing immediate feedback, we are told that 70% answered correctly.
Thus; p = 70% = 0.7
Similar to A above and using online binomial probability calculator, we have;
P(11) = 0.0654
P(12) = 0.1144
P(13) = 0.1643
P(14) = 0.1916
P(15) = 0.1789
P(16) = 0.1304
P(17) = 0.0716
P(18) = 0.0278
P(19) = 0.0068
P(20) = 0.0008
Thus;
P(k > 10) = 0.0654 + 0.1144 + 0.1643 + 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.952 = 95.2%