Download Because they are so easy to correct, multiple choice

Because they are so easy to correct, multiple choice questions are commonly used for
class test, SAT test, MCAT test for medical schools and many other circumstances. The
table listed describes the probability distribution for the number of correct responses
when someone makes random guesses for 10 multiple choice questions on a SAT test.
Each question has 5 possible answers (a,b,c,d,e) one of which is correct. Assume that
random guesses are made for each of the 10 questions. Please refer to the table for
questions 2-8
X
0
1
2
3
4
5
6
7
8
9
10
P (x)
0.107
0.268
0.302
0.201
0.088
0.026
0.006
0.001
0+
0+
0+
2) Find the mean number of correct responses?
Mean = Number of guess / number of options = 10 / 5 = 2
(note: it can also be obtained by summing the product of X and P(x), see EXCEL.)
3) Find the median number of correct response?
Theoretically, the median is the number in the middle of the range [0, 10], thus, it is
number 5.
4) Find the Standard Deviation for the number of correct responses when random guesses
are made for the 10 questions?
The formula of Variance is Var = Σ [(X-Mean)2 * P(X)] = 1.604
Then standard deviation = SQRT(Var) = 1.266
(SQRT is square root.)
5) What is the probability that someone gets at least half of the questions correct?
P(X> = 5) = P(5) + P(6) + … + P(10) = 0.033
6) When someone makes guesses for all 10 answers what is the expected number of
correct answers?
E(X) = Mean of X = 2
7) What is the probability of getting a least 1 answer correct?
P(X> = 1) = 1 - P(0) = 1 - 0.107 = 0.893
8) If someone gets at least 1 answer correct does that mean that this person knows
something about the subject matter being tested?
According to part 7, the probability of making at least 1 correct answer is as high as
0.893, while the person guess on all the 10 questions. Therefore, we cannot derive that
the person knows anything about the subject.
9) The television show Grey’s Anatomy has a 15% market share when its being broadcast
(1.e. 15% of TV sets are tuned into Grey’s Anatomy based upon data from Neilsen Media
Research) A special focus group consist of 12 randomly selected households (each with
one TV set in use during the time Grey’s Anatomy broadcast) Apply this to 10–14
10) What is the expected number of sets tuned into Greys?
E(X) = N * P(X) = 12 * 15% = 1.8
11) In such groups of 12 whish is the mean number of sets tuned into Grey’s?
Mean = E(X) = 1.8
12) In such groups of 12 what is he standard deviaton for the number of sets tuned in?
SD = SQRT[P*(1-P)/N] = SQRT[0.15*(1-0.15)/12] = 0.103
13) For such a group of 12 find the probability that exactly 3 sets are tuned in?
The probability of getting X sets in N samples is:
P(X)=N!/X!(N-X)! * p^x (1-p)^ (N-X)
When X = 3, N = 12, p =0.15, we have
P(3)=12!/3!(12-3)! * 0.15^3 * (1-0.15)^ (12-3) = 0.172
14) For such a group of 12 would it be unusual to find that no sets are tuned into Greys?
Why or why not?
When X = 0, N = 12, p =0.15, we have
P(0)=12!/0!(12-0)! * 0.15^0 * (1-0.15)^ (12-0) = 0.142
Since the probability of no sets turned in is as high as 14.2%, it is not unusual that no sets
are tuned in.
In the year 2001 The New York State Health Department reported a 10% rate of IV virus
infections for the “at risk” population in Westchester County - therefore an intensive
education program was conducted to lower that 10% rate. After running the program for
5 years a follow up study of 150 “at risk” individuals were surveyed. Apply this to
questions 16 and 17
16) Assuming that the program has no effect. Find the mean and standard deviation for
the number of HIV cases in the group of 150 people?
When the rate is still 10%, p = 0.1.
Mean = N * P(X) = 150 * 10% = 15
Variance = N * P(1-P) = 150 * 0.1 * (1-0.1) = 13.5
standard deviation = SQRT(Variance) = 3.67
17) Among the 150 people in the follow-up study 8% (or 12 people) tested positive for
the HIV virus. If the program has effect is the rate unusually low? Does this result
suggest that the program is effective?
We compute t-stat by
T = (12 - Mean) / (SD / sqrt(N)) = (12 - 15) / (3.67 / sqrt(150)) = -10.01
Since the absolute value of t-stat is 10, higher than the critical value (usually 2.0), we
have to say that the rate is significantly low and the program is statistically significantly
effective.
Inability to get along with others is the reason cited in 17% of employee firings in 2002.
Concerned about her company’s working conditions the personnel manager at the
Westchester Finance Company plans to investigate five employee firings that occurred
recently (apply this to questions 19 and 20)
19) Assuming that the 17% rate applies find the probability that among those five
employees the number fired because of an inability to get along with others is at least
four.
The probability of getting 4 in 5 samples is:
P(4)=5!/4!(5-4)! * 0.17^4 * (1-0.17)^ (5-4) = 0.00347
The probability of getting 5 in 5 samples is:
P(5)=5!/5!(5-5)! * 0.17^5 * (1-0.17)^ (5-5) = 0.00014
Thus, P(X> = 4) = P(4) + P(5) = 0.0036
20) If the personnel manager actually does find that at least four of the firings are due to
an inability to get along with others, does this company appear to be very different from
the typical companies? Why or Why not?
Since the probability of at least four of the firings due to an inability to get along with
other is as low as 0.0036 (which means less than 1% of all the companies have such a
case), it is very different from other companies.