Wednesday, October 20, 2010

Answers to Statistics and Probability Part 1: Mechanical Engineer Board Exam Review

Here's the solution and answers to the ME board questions in Mathematics (Statistics and Probability)

1. In how many ways can you arrange a group of 5 girls and 3 boys in 7 vacant chairs?
The total number of persons to be seated is 8 and the number of chairs is only 7.
When words denoting “arrangement” are used, it most likely pertains to permutations.
So, the formula to be used is permutation of 8 taking 7 at a time.
n = 8, r = 7, nPr, then use your calculator or this formula
nPr = n!/(n-r)! = 8!(8-1)!=40320 permutations or arrangements
A. 40320

2. How many 3-digit numbers can you make out of the numbers 1 to 5 without repetition?
Since the number varies as you rearrange the digits (example 1342 is different from 1423), then this problem is about permutation.
So, the formula is the permutation of 5 numbers taken 3 at a time.
n = 5, r = 3, nPr; use your calculator or this formula: nPr = n!/(n-r)!
nPr = 5!/(5-3)!=60 permutations
C. 60

3. There are 2 white, 3 red, and 4 blue balls inside a basket. If three balls are drawn randomly in succession without replacement, what is the probability that the first ball is white, and the next two balls are blue?
The probability that the first ball is white is Pw = 2/(2+3+4)=2/9
The probability that the second ball is blue is
Pb1=4/8 =1/2; we changed the denominator since the total number of balls has been reduced.
The probability that the last ball drawn is blue again is
Pb2 = 3/7; we reduced the total to 7 and the number of available blue is down to 3.
The probability that all this happen in correct succession is:
Pwbb = (2/9)(1/2)(3/7)=1/21
D. 1/21

4. What is the mode of the following numbers: 54, 45, 75, 60, 65, 65, 60, and 57?
Solution: the mode of a given set of numbers is the most frequent number; in this case those numbers are 65 and 65, both occurring twice.
D. 60 and 65

5. From the given numbers on question number 4, what is the median?
The media is the number that divides the upper and the lower half of the samples.
To find the media, arrange the data in order either increasing or decreasing
In this case: 45 54 57 60 60 65 65 75;
Since the number of samples is 8 (even), we have to get the average of the two middle samples
Median = (60+60)/2 = 60
B. 60

6. From the given numbers of question number 4, what is the variance?
Solution: variance is the measure of how much does the samples deviate from the mean or average. Variance (σ^2) is given by the formula σ^2= ∑[(x-ave)^2/(n-1)] or use your calculator
Average = 60.125
σ^2= ∑[(x-ave)^2/(n-1)] = 77.84
A. 77.84

7. Seven boys are to be seated around a circular table. How many arrangements can be made?
Since no indication as to the number of seats available, let’s assume that it is equal to the number of boys, 7
Again, this is a permutation problem, but this time, in circular arrangement
The formula for circular permutation is given by nPr (cyclic) = (n-1)!
So, nPr (cyc) = (7-1)! = 720
D. 720

8. In how many ways can you arrange 3 boys and 4 girls in a 7-seater bench supposing that the four girls want to be seated together?
Since one group wants to be together, then we will use conditional permutation techniques.
In this case, because the girls can’t be separated, let’s treat the girls group as one person or block.
The permutation of the boys and the group of girls therefore has the total number of n= 4, that is 3 boys plus 1 girls block.
The number of arrangement possible is given by nPr = 4P4=24
Remember though, that the girls can be rearranged as long as they are together, therefore, the total permutation inside the girls block is nPr = 4P4=24.
Combining the two permutations we have nPr (total) = (24) (24) = 576
C. 576

9. The probability that you will arrive late is 35% and the probability that you will scolded by your boss is 15%. What is the probability that you will be both late and scolded by your boss?
The problem doesn’t mention that the cause of being scolded is caused by you being late. With that we can treat those two events as independent from each other.
The probability that the two independent events will happen can be obtained by getting the product of each probabilities.
Pa&b = (Pa)(Pb) = (0.35)(0.15)=0.0525 = 5.25%
A. 5.25%

10. From question number 9, what is the probability that you will either be late or scolded by your boss?
This is an OR condition (the condition is valid if only when one of the two conditions happened) and is not mutually exclusive (meaning the two outcomes may happen at the same time: being late and being scolded)
The formula to be used is: Paorb = Pa+Pb – Pa&b
Paorb = 0.35+0.15-(0.35)(0.15)=0.4475
Paorb = 44.75%
C. 44.75%

Link to the questions.

Note: Writing solution to a math problem in a webpage like these is extremely difficult so I expect that some of you may be confused by the way I wrote the solution. If you have questions and clarifications, don't hesitate to send me a message at or post a comment below.

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