In a community, 5 people from Koc group, 6 people from Sabanci group and 4 peopl
ID: 3292228 • Letter: I
Question
In a community, 5 people from Koc group, 6 people from Sabanci group and 4 people from Independent group exist as representatives. One, who does not know all these representatives, creates a random committee with 3 people in total from these representatives. With respect to this, a) What is the probability of being represented in the committee with one person from each group? b) What is the probability of creating the committee from the representatives of the same group? c) What is the probability of creating the committee with 2 people from any of these three groups and 1 person from another group? Based on information from MRI Network, some job applicants are required to have several interviews before a decision is made. The number of required interviews and the corresponding probabilities are: 1 (0.09): 2 (0.31): 3 (0.37): 4 (0.12): 5 (0.05): 6 (0.05). a) Does the given information describe a probability distribution? Explain b) Assuming that a probability distribution is described, find its mean and standard deviation? c) Use the range rule of thumb to identify the range of values for usual numbers of interviews d) Is it unusual to have a decision after just one interview? Explain. a) A simple random sample of a credit rating scores is listed below. As of this writing, the mean credit rating score was reported to be 678. Based on these results, is a credit rating score of 500 unusual? Why or why not? b) Construct a box-plot and include the values of the 5-number summary. The National Highway Traffic Safety Administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). Find the a) mean b) median c) mode, and d) midrange for the given data. According to the safety requirement, the hic measurement should be less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement? a) Identify the class width, class midpoints and class boundaries for the given frequency distributions. b) Construct one table that includes relative frequencies based on frequency distributions for non-filtered and filtered cigarettes, and then compare the amounts of tar in both types of cigarettes. Do the cigarette filters appear to be effective? c) Construct the cumulative frequency distributions that correspond to the frequency distributions in the question indicated.Explanation / Answer
Solution
Back-up Theory
1. Number of ways of selecting any r out of n things is given by nCr = (n!)/{(r!)(n - r)!}, where n! = n(n - 1)(n - 2)…. 1
2. Probability of an event, E, is given by P(E) = n/N where n = number of possibilities (outcomes) that lead to E and N = total number of all possibilities (outcomes)
3. In the given question, 3 people are selected out of 15 people of 3 different groups. So, total number of possibilities = 15C3 = (15!)/{(3!)(12)!} = 455.
Hence by (2) above, the denominator for all the probabilities would be 455.
Q1Part (a)
Number of ways of selecting 3 people out of 15 such that there is one person from each group = (5C1)(6C1)(4C1) = 5 x 6 x 4 = 120.
So, required probability = 120/455 = 24/91 = 0.2637 ANSWER
Q1Part (b)
Number of ways of selecting 3 people out of 15 such that all three belong to the same group
= (5C3) + (6C3) + (4C3) = 10 + 20 + 4 = 34.
So, required probability = 34/455 = 0.0747 ANSWER
Q1Part (c)
Two from one group and one from a different group is made up of the following possibilities
Case
Number of persons from group
# of possibilities
(formula)
# of possibilities
(number)
Koc
Sabanci
Independent
1
2
1
0
(5C2)(6C1)(4C0)
60
2
2
0
1
(5C1)(6C0)(4C1)
40
3
1
2
0
(5C1)(6C2)(4C0)
50
4
1
0
2
(5C1)(6C0)(4C2)
30
5
0
1
2
(5C0)(6C1)(4C2)
36
6
0
2
1
(5C0)(6C2)(4C1)
60
TOTAL
276
So, required probability = 276/455 = 0.6066 ANSWER
Case
Number of persons from group
# of possibilities
(formula)
# of possibilities
(number)
Koc
Sabanci
Independent
1
2
1
0
(5C2)(6C1)(4C0)
60
2
2
0
1
(5C1)(6C0)(4C1)
40
3
1
2
0
(5C1)(6C2)(4C0)
50
4
1
0
2
(5C1)(6C0)(4C2)
30
5
0
1
2
(5C0)(6C1)(4C2)
36
6
0
2
1
(5C0)(6C2)(4C1)
60
TOTAL
276