The variable x takes on integer values between, 1 and 7 inclusive, as

Question

FREQUENCY DISTRIBUTION OF THE VARIABLE X
X	Freequency
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 3/2?

(A) 8/15
(B) 4/7
(C) 4/5
(D) 6/7
(E) 8/7

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Bunuel · Accepted Answer

According to the table, we have the following list of 3 + 1 + 3 + 1 + 3 + 1 + 3 = 15 numbers with the mean of 4: {1, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 7} (so, three 1's, one 2, three 3's, ...). We need to find the probability that |4 - x| > \frac{3}{2} ( the absolute value of the difference between the mean and x is |4 - x| ). |4 - x| > \frac{3}{2} ; 4 - x > \frac{3}{2} or 4 - x< -\frac{3}{2} ; x>5.5 or x<2.5 Thus, the question simply asks to find the probability that a randomly chosen x from {1, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 7} will be more than 5.5 (6, or 7) or be less than 2.5 (2 or 1). P(x = 1, 2, 6, \ or \ 7) = \frac{3+1+1+3}{15}=\frac{8}{15} Answer: A. Hope it's clear.

MacFauz · Answer

The set is
1,1,1,2,3,3,3,4,5,5,5,6,7,7,7

The question basically asks for the probability of choosing 1,2,6 or 7.

= 8/15

bhavinshah5685 · Answer

Hi Can you please explain in detail?

how u got the values of set?

MacFauz · Answer

Looking at the table.,

The column on the left shows the value of X and the column on the right shows the number of times the value appears. The question statement is just beating around the bush a lot to ask a simple question. What is the probability that 4 - X > 1.5. For such a case, X has to be 1,2,6 or 7.

Zarrolou · Answer

x has values between 1 and 7 so  |4-x|>\frac{3}{2} =

|4-1|=3>3/2  
  |4-2|=2>3/2  
 |4-3|=1>3/2
|4-4|=0>3/2
|4-5|=1>3/2 
  |4-6|=2>3/2  
  |4-7|=3>3/2

The value that respect |4-x|>3/2 are 1,2,6,7 that have a frequency of 3,1,1,3 so  3+1+1+3  = total frequencies of "right" events, and this will be our  numerator .
Our  denominator  will be the tot number of combinations =  3+1+3+1+3+1+3= 15. 
Answer  \frac{8}{15}

knightofdelta · Answer

How can the probability of doing something be 3/2?

How can the probability of doing something be greater than 1?

I don't understand what y'all are calculating.

fozzzy · Answer

The final answer is below 1

8/15 = 0.5333

Woody19 · Answer

Q - 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 3/2?

i.e |x - 4 | > 3 / 2

according to absolute value rules above equation becomes.

x-4 > 3 / 2 and x-4 < -3/2

solving both - we get : x > 5.5 and x < 2.5.

from the table we have - 6 ,7 ,7 7 above 5.5 and 2 , 1 ,1 ,1 below 2.5 . total - 8 occurrences.

probability = number of occurrences / total occurrences

P = 8 / 15

Ans - 8 / 15

Thanks.

Sushen · Answer

The answer is A. I believe based on these answers, some are still confused. Here is a more complete explanation:

Interpret/Translate the Question so that you can answer it.
What is the probability: [chances of the outcome you are interested in ÷ all of the data points]
Absolute value: |data-mⅇan| is the distance between your random point and the mean.
It makes the distance equal whether it is above or below the mean.
The mean is 4
What is the probability that a random point is greater than or less than a distance of 3/2 from the mean?
i.e. x<4-3/2 OR x>4+3/2 ? x<2.5 OR x>5.5
In probability I must add these to include all data points that I am interested in.

4 data points are <2.5, 4 data points are >5.5.
There are 8 data points that meet our criteria.
There are 15 data points.
The probability is 8/15.

CrushTHYGMAT · Answer

Hi  KarishmaB ,  avigutman ,  MartyTargetTestPrep ,  CrackVerbal  I marked option B. The reason is I looked at this question in the following way:

We can make 7 sets (4-1) (4-2) (4-3) (4-4) (4-5) (4-6) (4-7). Out of that, we are only interested in set 1, 2, 6, and 7. So, we get 4/7. I didn't count (4-1) as three separate sets since there are three 1s.

Can you please help me understand the fault in my logic? Can we not count three (4-1) set as one because there are actually three different values 1, 1, 1 that we have?

Thank you.

KarishmaB · Answer

Think about it: 
Say you have 3 numbers 1, 1 and 10. 
What is the probability that the number you pick randomly from these three will be less than 5? Is it 1/2? But that would be the probability of picking a number less than 5 if you had just two numbers 1 and 10. You actually have 3 numbers and 2 of them are less than 5, even if they are equal. 
So the probability of picking a number less than 5 is 2/3.

We use the same logic here. The numbers are 1, 1, 1, 2, 3, 3, 3 ... etc.

CrushTHYGMAT · Answer

Thank you so much,  KarishmaB . It makes perfect sense now.

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