Psych Stats - Help with Review Q'z

#1
I just received a stats review sheet with tons of questions which I am suppose to do as exam review for my psych stats course and I have a few questions that I am having some problems with

I already have the answers to these problems, so that’s not what I am looking for… with stats, I usually just don’t know which concepts and formulas to apply when and where so that’s what I need help with

1)A sample of 200 people were given a test. The distribution of the test scores was mound-shaped and symmetrical with a mean of 100. One person, whose test score was 125, was found to be at the 84th percentile.

a)Approximately how many people obtained scores between 65 and 85?

Using Chebyshev’s theorem, a score of 65 corresponds to a standard deviation of 1.4 and a score of 85 corresponds to a standard deviation of 0.6 but I don’t know where to go from there … I believe my prof told me the answer to this question was 38 people

b)Suppose the distribution was not mound-shaped, but rather skewed. At least how many people would be expected to obtain scores between 65 and 135?

This is how I did the problem but my answer (98.97) is a little off from the original answer (98) so I don’t know where I am going wrong with this:
Z1 = 65-100/25 = -1.4 and Z2 = 135-100/25 = 1.4 so value of k = 1.4
Using Chebyshev’s theorem, 1 – (1/1.4^2) = 1-0.510 = 0.48979 x 100 = 48.97 + 50 = 98.97


2)Charles Darwin discovered a strange species of shrubs on a remote island and recorded their heights in inches as follows: “I measures 800 plants. The distribution of their heights is mound-shaped and symmetrical and has a variance of 9. One plant, whose height was 4 inches, falls at the 33rd percentile of the distribution.”

a)Approx how many of the 800 plants in this sample are between 8.5 and 10.5 inches tall?

I realize that this question is similar to the one above, but how can I use Chebyshev’s theorem without knowing the mean? The answer to this question is suppose to be 72 plants

b)Approx what percentage of the sample of 800 plants would be expected to have a Z-score of 1.28? (Assuming plant heights can be positive or negative)

Answer is 87.78%

4)A general is planning to invade towns A, B and C and has 20 soldiers at his disposal (6 officers and 14 privates). After some thought, the general decides to select 12 soldiers to carry out the invasion and to keep the remaining 8 and himself, behind to protect the command post. If the general takes 12 selected soldiers (3 officers and 9 privates), and randomly selects 4 soldiers without replacement, to invade each towns A, B and C, what is the probability that exactly 1 officer ends up being sent to each town?

5)A multiple choice test contains 12 questions, 8 of which have 4 answers each to choose from and 4 of which have 5 answers to choose from. If a student randomly guesses all of his answers, what is the probability that he will get exactly 2 of the 4 answer questions correct and at least 3 of the 5 answer questions correct?

Out of the 8-four answer questions, the student gets 2 of them = (8C2)
Out of the 4-five answer questions, the student gets 3 of them = (4C3)
So probability = (8C2) x (4C3) / [12!/8!4!] 28 x 4 / 495 = 0.22626 was my answer but the correct answer is suppose to be 0.2552


6)A standard deck of 52 cards consist of 4 suits, each of which contains 13 cards. A player selects 3 cards at random, without replacement, and is interested in knowing the mean and standard deviation of X: the number of diamonds in his selection. Compute the answers for him.

And this question I have no idea how to do but I was told if I did it correctly the mean would come out to be -0.75 and the SD would be 0.736
 
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JohnM

TS Contributor
#2
I'll respond for a couple of these:

5) Your answer only takes into account exactly 2 of 4 and exactly 3 of 5, and you need to consider "at least" 3 of the 5 (i.e., 4 of 5 and 5 of 5).
That's why your answer is slightly off.

6) This is a discrete probability distribution. If you draw 3 cards at random, without replacement, the number of possible diamonds (x) could be 0,1,2, or 3. You need to find the probability of each.

For x=0, the probability is (39/52) * (38/51) * (37/50).
For x=3, the probability is (13/52) * (12/51) * (11/50).
For x=1 or x=2, the method is basically the same, except you need to consider on which of the 3 draws a diamond occurs, and it can occur a few different ways....
 
#3
for the multiple choice question

would i be correct in stating the following:

The probability of getting exactly two of the eight four-option questions is:
[8C2(.25)2(.75)6] = 0.31146

The probability of getting at least three of the four five-option questions is:
[ 4C3(.2)3(.8)1] = 0.0256

The probability of getting at four of the four five-option questions is:
[ 4C4(.2)4(.8)0]

therefore, [8C2(.25)2(.75)6][ 4C3(.2)3(.8)1+ 4C4(.2)4(.8)0] = 0.008471 which does not match with the given answer of 0.2552
 

JohnM

TS Contributor
#4
exactly 2 out of 8 four-choice questions:

8C2 * (.25)^2 * (.75)^6 = .3115

at least 3 out of 4 five-choice questions:

[4C3 * (.2)^3 * (.8)] + [4C4 * (.2)^4] = .0272

add them up: .3115 + .0272 = .3387