Entries Tagged as '2: stats discrete math'

Sample Test Item 58 (2003)

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This question is testing your ability to calculate permutations.

They are asking for the number of permutations of n things taken n at a time.

Specifically, the are asking for number of ways you can arrange 6 students in EVERY order.

Mathematician use factorial notation to represent the number of possible permutations for situations like this.

There are 6 seats and 6 students. Six seats taken at six students at a time.

6P6 = 6!

6! = (6)(5)(4)(3)(2)(1)

As you can see by multiplying the first 3 factors, the only possible answer is Choice A.

Tags: 2: stats discrete math, 5: structure logic, sample test 2003 by admin
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Sample Test Item 57 (2003)

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Choice D is wrong. The points on this graph are not random, so there appears to be a relationship between te test scores and the days absent.

Choice A is also wrong. This makes no sense.

There are only two choices left: Choice B and Choice C.

We see there is a downward (negative) slope that shows a trend of test scores as they are related to days absent.

It is safe to say that as students missed more days of school, their test score decreased.

Choice C is the correct answer.

Tags: 2: stats discrete math, sample test 2003 by admin
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Sample Test Item 60 (2003)

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this is a useless problem.
skip it.

i have never seen this in real life or in business on a daily basis.

IGNORE this ridiculous question. ^_^ …

Tags: 2: stats discrete math, sample test 2003 by admin
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Sample Test Item 48 (2003)

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Why is donald trump still rich?

Because he understands this chart!

Choice D is eliminated because it outside of the time range of the chart.

Choice B is wrong because growth of money is exponential.

Only choice C provides a good summary of the information in the graph.

Choice C is the only answer.

Tags: 2: stats discrete math, sample test 2003 by admin
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Sample Test Item 36 (2003)

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three cubes with sides number 1 through 6? hmm… apparently these test makers have never heard of a casino or gambling or Las Vegas.

i believe they are called DICE!!

the chance of landing on 1 face is (1/6).

the chance of all three dice landing on the same face labeled 3 is (1/6) times (1/6) times (1/6).

P3 = (1/6)3

the chance of all three dice landing on the same face labeled 4 is (1/6) times (1/6) times (1/6).

P4 = (1/6)3

total probablity is:

Ptotal = P3 + P4 = (1/6)3 + (1/6)3
Ptotal = 2 * (1/6)3

choice B is the correct answer

Tags: 2: stats discrete math, sample test 2003 by admin
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Sample Test Item 22 (2003)

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The problem starts off with useless babble about food, which has NOTHING to do with calculation of the median. The question is “what is the median of the data set?”

First re-order the data set. Be careful not to miss a number.

So, you will have: [0.3, 0.3, 0.5, 0.6, 0.9, 1.1, 1.1, 1.1, 4.0, 5.1].

Becuase we have an even number of data numbers, we take the 5th and 6th numbers and average them to find the actual median.

The median is 1.0, which is Choice B.

Tags: 2: stats discrete math, sample test 2003 by admin
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