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A number is randomly chosen from a list of 10 consecutive po
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30 Jul 2018, 10:37
30 Jul 2018, 10:37
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A number is randomly chosen from a list of 10 consecutive positive integers. What is the probability that the number selected is greater than the average (arithmetic mean) of all 10
integers?
A. \(\frac{3}{10}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{10}\)
E. \(\frac{4}{5}\)
integers?
A. \(\frac{3}{10}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{10}\)
E. \(\frac{4}{5}\)
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Re: A number is randomly chosen from a list of 10 consecutive po
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02 Aug 2018, 16:11
02 Aug 2018, 16:11
1
sandy wrote:
A number is randomly chosen from a list of 10 consecutive positive integers. What is the probability that the number selected is greater than the average (arithmetic mean) of all 10
integers?
A. \(\frac{3}{10}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{10}\)
E. \(\frac{4}{5}\)
integers?
A. \(\frac{3}{10}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{10}\)
E. \(\frac{4}{5}\)
We want to determine P(the number selected is greater than the average all 10 integers)
The average of all 10 integers = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)/10 = 55/10 = 5.5
So, we really want to determine P(selected number is greater than 5.5)
Let's see how many of the 10 integers are greater than 5.5
They are: 6, 7, 8, 9, and 10
There are 5 integers in total
So, P(selected number is greater than 5.5) = 5/10 = 1/2
Answer: C
Cheers,
Brent
Re: A number is randomly chosen from a list of 10 consecutive po
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07 Aug 2018, 05:58
07 Aug 2018, 05:58
Expert Reply
\(Explanation\)
In a list of 10 consecutive integers, the mean is the average of the 5th and 6th numbers.
Therefore, the 6th through 10th integers (five total integers) is greater than the mean. Since probability is determined by the number of desired items divided by the total number of choices, the probability that the number chosen is greater than the average of all 10 integers is \(\frac{5}{10} = \frac{1}{2}\).
Another approach to this problem is to create a set of 10 consecutive integers; the easiest such list contains the numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The mean is one-half the sum of the first element plus the last element, or \(\frac{10+1}{2}= 5.5\).
Therefore, there are 5 numbers greater than the mean in the list: 6, 7, 8, 9 and 10. Again, the probability of choosing a number greater than the average of all 10 integers is \(\frac{5}{10} = \frac{1}{2}\).
In a list of 10 consecutive integers, the mean is the average of the 5th and 6th numbers.
Therefore, the 6th through 10th integers (five total integers) is greater than the mean. Since probability is determined by the number of desired items divided by the total number of choices, the probability that the number chosen is greater than the average of all 10 integers is \(\frac{5}{10} = \frac{1}{2}\).
Another approach to this problem is to create a set of 10 consecutive integers; the easiest such list contains the numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The mean is one-half the sum of the first element plus the last element, or \(\frac{10+1}{2}= 5.5\).
Therefore, there are 5 numbers greater than the mean in the list: 6, 7, 8, 9 and 10. Again, the probability of choosing a number greater than the average of all 10 integers is \(\frac{5}{10} = \frac{1}{2}\).
