\(x\) frequency
1 3
2 1
3 3
4 1
5 3
6 1
7 3
The variable \(x\) takes on integer values between \(1\) and \(7\) inclusive as shown above. What is the probability that the absolute value of the difference between the mean of the distribution which is \(4\) and a randomly chosen value of \(x\) will be greater than \(\dfrac32?\)
A) \(\dfrac8{15}\)
B) \(\dfrac47\)
C) \(\dfrac45\)
D) \(\dfrac67\)
E) \(\dfrac87\)
Answer: A
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The variable \(x\) takes on integer values between \(1\) and \(7\) inclusive as shown above. What is the probability
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Solution:M7MBA wrote: ↑Thu Sep 17, 2020 1:18 am\(x\) frequency
1 3
2 1
3 3
4 1
5 3
6 1
7 3
The variable \(x\) takes on integer values between \(1\) and \(7\) inclusive as shown above. What is the probability that the absolute value of the difference between the mean of the distribution which is \(4\) and a randomly chosen value of \(x\) will be greater than \(\dfrac32?\)
A) \(\dfrac8{15}\)
B) \(\dfrac47\)
C) \(\dfrac45\)
D) \(\dfrac67\)
E) \(\dfrac87\)
Answer: A
The only values of x that give us |x - 4| > 3/2 are 1, 2, 6, and 7. Since we have a total of 3 + 1 + 3 + 1 + 3 + 1 + 3 = 15 values in the list and a total of 3 + 1 + 1 + 3 = 8 values that can be x, the probability is 8/15.
Answer: A
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