numbers 1 to 20 ,random 5 numbers. probability that this random variable will be less than 7?

[lunarsrh]

Hey everyone. This problem was in my exam and I couldn't find my answer in multiple answer choices that were given. and I'm not sure if I'm doing it wrong or the choices were wrong.

From the numbers 1 to 20, we randomly select 5 numbers. If the random variable X represents the median of these selected numbers, what is the probability that this random variable will be less than 7?

I could really use some help. Thank you in advance. (On a side note, I don't speak English so I had to translate the problem. I hope it's still clear!)

 

[nasi112]

There are [MATH]136[/MATH] ways to have [MATH]3[/MATH] as the median.

There are [MATH]360[/MATH] ways to have [MATH]4[/MATH] as the median.

There are [MATH]630[/MATH] ways to have [MATH]5[/MATH] as the median.

There are [MATH]910[/MATH] ways to have [MATH]6[/MATH] as the median.

There are [MATH]15504[/MATH] ways to select [MATH]5[/MATH] numbers from [MATH]20[/MATH] numbers.

So, the probability to have a median less than [MATH]7[/MATH] is

[MATH]p(X < 7) = \frac{136 + 360 + 630 + 910}{15504} = \frac{509}{3876} = 0.13[/MATH]
Note: [MATH]1[/MATH] and [MATH]2[/MATH] will never be the median.

 

[pka]

Note: [MATH]1[/MATH] and [MATH]2[/MATH] will never be the median.

Note that you have assumed that the sample is without replacement.

 

[Deleted member 4993]

There are [MATH]136[/MATH] ways to have [MATH]3[/MATH] as the median.

There are [MATH]360[/MATH] ways to have [MATH]4[/MATH] as the median.

There are [MATH]630[/MATH] ways to have [MATH]5[/MATH] as the median.

There are [MATH]910[/MATH] ways to have [MATH]6[/MATH] as the median.

There are [MATH]15504[/MATH] ways to select [MATH]5[/MATH] numbers from [MATH]20[/MATH] numbers.

So, the probability to have a median less than [MATH]7[/MATH] is

[MATH]p(X < 7) = \frac{136 + 360 + 630 + 910}{15504} = \frac{509}{3876} = 0.13[/MATH]
Note: [MATH]1[/MATH] and [MATH]2[/MATH] will never be the median.

@nasi112 - you had a nice chance to teach - instead you decided to write down total solution! You did "all these" before the OP showed a line of original effort.

 

[nasi112]

Note that you have assumed that the sample is without replacement.

Yup, it is probably without replacement or the student will have no chance to answer the question.

@nasi112 - you had a nice chance to teach - instead you decided to write down total solution! You did "all these" before the OP showed a line of original effort.

Don't worry Khan. The OP will be wondering how those numbers been found. Then, he/she will be asked to show how did he/she attack the problem. Then, we will provide him/her more information.

Those numbers are useless for the OP, unless he/she is in a Test now.

 

[Deleted member 4993]

Yup, it is probably without replacement or the student will have no chance to answer the question.
Don't worry Khan. The OP will be wondering how those numbers been found. Then, he/she will be asked to show how did he/she attack the problem. Then, we will provide him/her more information. Those numbers are useless for the OP, unless he/she is in a Test now.

But you need to guide the student - after s/he has shown some effort related to the problem. You did the same at "relative distance" problem - and provided the WRONG answer. In this thread you missed to state an important condition - non replacement.

 

[nasi112]

But you need to guide the student - after s/he has shown some effort related to the problem. You did the same at "relative distance" problem - and provided the WRONG answer. In this thread you missed to state an important condition - non replacement.

It is logical to assume without replacement. Even with replacement there is a nice solution, but I hope I am right about "WITHOUT".

And yeah, I did a mistake in relative distance, but I did not leave the thread. We have discussed about it, and the OP left happy.

Everyone makes mistakes, but you're right, I have to wait for the OP to show some efforts.

Next days, you will see a different nasi. Trust me.

AND easy on Girls, we are emotional.

 

[Dr.Peterson]

Hey everyone. This problem was in my exam and I couldn't find my answer in multiple answer choices that were given. and I'm not sure if I'm doing it wrong or the choices were wrong.

From the numbers 1 to 20, we randomly select 5 numbers. If the random variable X represents the median of these selected numbers, what is the probability that this random variable will be less than 7?

I could really use some help. Thank you in advance. (On a side note, I don't speak English so I had to translate the problem. I hope it's still clear!)

When you ask for help, you need to show the entire question, including the list of choices, in case perhaps the problem itself is wrong, or it clarifies what the problem means (e.g. selection with or without replacement).

You also need to show us your work, in case you actually got the right answer, or came very close to it but made a small mistake. That would save a lot of work in helping you.

 

[lunarsrh]

There are [MATH]136[/MATH] ways to have [MATH]3[/MATH] as the median.

There are [MATH]360[/MATH] ways to have [MATH]4[/MATH] as the median.

There are [MATH]630[/MATH] ways to have [MATH]5[/MATH] as the median.

There are [MATH]910[/MATH] ways to have [MATH]6[/MATH] as the median.

There are [MATH]15504[/MATH] ways to select [MATH]5[/MATH] numbers from [MATH]20[/MATH] numbers.

So, the probability to have a median less than [MATH]7[/MATH] is

[MATH]p(X < 7) = \frac{136 + 360 + 630 + 910}{15504} = \frac{509}{3876} = 0.13[/MATH]
Note: [MATH]1[/MATH] and [MATH]2[/MATH] will never be the median.

Yes!! then my answer was correct

[MATH]\sum_{i=3}^{5} \frac{ \binom{6}{i} \binom{14}{5-i} }{ \binom{20}{5} }[/MATH]
this is what I got which also equals to 2036/15504 = 0.13
I was just so confused because I couldn't find the answer in given choices and I tried to convince my professor that the right answer was on the question but he didn't really paid attention.

Thank you so much

 

[lunarsrh]

When you ask for help, you need to show the entire question, including the list of choices, in case perhaps the problem itself is wrong, or it clarifies what the problem means (e.g. selection with or without replacement).

You also need to show us your work, in case you actually got the right answer, or came very close to it but made a small mistake. That would save a lot of work in helping you.

That was the entire question. I guess they didn't mention replacement. I just didn't show the choices which now I know for sure were wrong.
and yes you're right I could show that I made some effort. It was my first time here so I didn't really know what to do. Thank you for your help. I do better next time!

 

[lunarsrh]

Yup, it is probably without replacement or the student will have no chance to answer the question.



Don't worry Khan. The OP will be wondering how those numbers been found. Then, he/she will be asked to show how did he/she attack the problem. Then, we will provide him/her more information.

Those numbers are useless for the OP, unless he/she is in a Test now.

also don't worry I'm not in a test now. It's 3:22 A.M. here where I live! I had this test about 2 weeks ago and I was still pissed that I couldn't find the answer so I came here to ask and make sure I got it right.

 

[nasi112]

Yes!! then my answer was correct

[MATH]\sum_{i=3}^{5} \frac{ \binom{6}{i} \binom{14}{5-i} }{ \binom{20}{5} }[/MATH]
this is what I got which also equals to 2036/15504 = 0.13
I was just so confused because I couldn't find the answer in given choices and I tried to convince my professor that the right answer was on the question but he didn't really paid attention.

Thank you so much

Did you try to solve it with replacement? You might get an answer that was in the choices.

Or

Can you tell us what were the choices?

 

[Diamond909]

Help me with this please

 

[Dr.Peterson]

That was the entire question. I guess they didn't mention replacement. I just didn't show the choices which now I know for sure were wrong.
and yes you're right I could show that I made some effort. It was my first time here so I didn't really know what to do. Thank you for your help. I do better next time!

We absolutely do need to see the choices. The fact that you got the correct answer for the without-replacement case is not enough to conclude that your teacher was wrong (except in not specifying that). The null hypothesis is that the teacher is right.

 

[lunarsrh]

Did you try to solve it with replacement? You might get an answer that was in the choices.

Or

Can you tell us what were the choices?

The choices did not make sense at all. It was an online test so I guess they got mixed up with another question or something. But maybe I'm missing something

[MATH]\sum_{i=2}^{15}\frac{(\begin{matrix}i\\2\\\end{matrix})(\begin{matrix}19-i\\2\\\end{matrix})}{(\begin{matrix}20\\3\\\end{matrix})}[/MATH]
[MATH]\sum_{i=2}^{5}\frac{(\begin{matrix}i\\2\\\end{matrix})(\begin{matrix}19-i\\2\\\end{matrix})}{(\begin{matrix}20\\3\\\end{matrix})}[/MATH]
These were the choices.

 

[Dr.Peterson]

Help me with this please

This does not belong in this thread. Please start a new thread, and also state the problem clearly -- I can't quite tell what the problem is, and you've given no information as to where you need help. Follow the guidelines.

 

[Dr.Peterson]

The choices did not make sense at all. It was an online test so I guess they got mixed up with another question or something. But maybe I'm missing something

[MATH]\sum_{i=2}^{15}\frac{(\begin{matrix}i\\2\\\end{matrix})(\begin{matrix}19-i\\2\\\end{matrix})}{(\begin{matrix}20\\3\\\end{matrix})}[/MATH]
[MATH]\sum_{i=2}^{5}\frac{(\begin{matrix}i\\2\\\end{matrix})(\begin{matrix}19-i\\2\\\end{matrix})}{(\begin{matrix}20\\3\\\end{matrix})}[/MATH]
These were the choices.

Only two choices? And not numbers but formulas? That's very odd!

In particular, there are multiple ways to find a solution, which might not correspond to the same formula. Also, the two formulas differ only in the upper limit, and "i choose 2" makes no sense.

Did your professor even look at the problem?

 

[Cubist]

... "i choose 2" makes no sense.

The "i choose 2" could be explained if (i+1) is the median point. Then, i choose 2 would be the ways of choosing two numbers that are less than the median. And (19-i) choose 2 would be the ways of choosing two numbers greater than the median (without replacement).

But the denominator is very wrong, it ought to be 20 choose 5

 

[lunarsrh]

Only two choices? And not numbers but formulas? That's very odd!

In particular, there are multiple ways to find a solution, which might not correspond to the same formula. Also, the two formulas differ only in the upper limit, and "i choose 2" makes no sense.

Did your professor even look at the problem?

It was actually 4 choices because the question had another part asking for the support of the random variable. But only these two options for this part of the question.



Like this.
And honestly no I don't think he looked at the problem. It was an online test and I believe the questions were generated and he only selected them. I sent him an email and he said the answer is the fourth one and I'm not sure if he really checked it himself or looked at the answer key to the test.

 

[lunarsrh]

The "i choose 2" could be explained if (i+1) is the median point. Then, i choose 2 would be the ways of choosing two numbers that are less than the median. And (19-i) choose 2 would be the ways of choosing two numbers greater than the median (without replacement).

But the denominator is very wrong, it ought to be 20 choose 5

Ahhh yes I get it now. the only problem now is the 3 in the denominator.
Thank you so much!