Binomial Probability?

prettylittlepixels

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57% of all Hobbits reside in the SAME province in which they were born. A sample of 19 Hobbits is randomly selected from the population.

How many of the 19 Hobbits in the sample could be expected to be living in a province DIFFERENT from that in which they were born.

.43*19= 8.17, must be a whole number, so 8.


What is the standard deviation of this binomial probability function?

[FONT=MathJax_Math]σ[FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size1]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]19[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]0.43[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]0.57[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size1]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2.16[/FONT][/FONT]



Using the binomial probability function:
P(x) = (binomial coefficient)(p)^x(q)^(n-x),
NOT THE NORMAL APPROXIMATION TO THE BINOMIAL, what is the probability that more than 7 of the 19 Hobbits reside in a province DIFFERENT from that in which they were born?
REMEMBER: p and x must be consistent with each other (they must both represent "success").

[P(0)+P(1)... P(7)] = 0.3564649998

1- 0.356464998= .6435350002

Am I done here with it rounded to 0.6435?


Would it be appropriate to use the normal approximation of the binomial to answer #3 above? Answer 1, 2, 3, or 4 below.
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1. Yes, because np and nq are both greater than or equal to 5.
2. No, because either np or nq is less than 5.
3. No, because n is less than 20.
4. Yes, because n is greater than 100.


1.Yes, because np and nq are both greater than or equal to 5. Because np= .43*19=8.17 and nq= .57*19= 10.83

USING THE NORMAL APPROXIMATION TO THE BINOMIAL (whether or not it is appropriate, use it), what is the probability that more than 7 of the 19 Hobbits reside in a province DIFFERENT from that in which they were born?
REMEMBER: p and x must be consistent with each other (they must both represent "success").
REMEMBER: The number of Hobbits is a discrete binomial so you must perform the continuity correction to approximate the probability using the normal distribution.


(7.5-8.17)/2.16= 0.3101851852
.31 on Z table is .1217
5-.1217= 4.8783

BUT BECAUSE THIS IS A PROBABILITY IT WOULD STILL BE .1217?


 
Last edited:
57% of all Hobbits reside in the SAME province in which they were born. A sample of 19 Hobbits is randomly selected from the population.

I think this question is really throwing me off. It's been some time since I've practiced this and I've looked through the book and can't find the formula I need to use. I know with the right formula, I can figure out all of the other components to the question. I have worked out the problem below, but with a shove in the right direction, I think I can get this. Thanks!

How many of the 19 Hobbits in the sample could be expected to be living in a province DIFFERENT from that in which they were born.

I know 57% reside in the SAME province so 43% will be from a DIFFERENT province than they were born.

Do I just multiply .43 by 19= 8.17. I know I'm missing something here because that is just too simple.
Simple, but correct. The expectation value of a binomial distribution is p*n = (0.43)*(19) = 8.17

Using the binomial probability function:

P(x) = (binomial coefficient)(p)^x(q)^(n-x),

what is the probability that more than 7 of the 19 Hobbits reside in a province DIFFERENT from that in which they were born?
I will re-write the binomial probability to look better, and to put in p and n:

\(\displaystyle \displaystyle P(x) = \binom{19}{19-x}\ (0.43)^x\ (0.57)^{19-x}\)

You are trying to find P(>7). You have two choices:

either ...P(>7) = P(8) + P(9) + . . . + P(19)

or .........P(>7) = 1 - P(<7) = 1 - [P(0) + P(1) + . . . + P(7)]

The first form requires you to calculate 12 probabilities, while the second "only" needs 8. I would use the second. This calculation gives you the exact probability, but takes a lot of calculation. Often we can (or should) use a "normal approximation" to the binomial distribution.

USING THE NORMAL APPROXIMATION TO THE BINOMIAL (whether or not it is appropriate, use it), what is the probability that more than 7 of the 19 Hobbits reside in a province DIFFERENT from that in which they were born?
You must have been given some empirical rules to guide when the normal approximation is good. I don't know what you were taught - look at your text or notes to see whether choice 1 (Yes) or choice 3 (No) fits the rules you have.

The standard deviation of a binomial also has a form you should memorize:

\(\displaystyle \sigma = \sqrt{n\ p\ (1-p)} = \sqrt{19(0.43)(0.57)} = 2.16 \)

Thus the normal distribution that is an approximation to this binomial has mean 8.17 and standard deviation 2.16.

REMEMBER: p and x must be consistent with each other (they must both represent "success").
REMEMBER: The number of Hobbits is a discrete binomial so you must perform the continuity correction to approximate the probability using the normal distribution.
That means that the z-value you use will correspond to 7.5 Hobbits, because any number greater than 7.5 will round to be >7. Find z and look in a table of the normal distribution to find the probability of >z.

Compare this result (the normal approximation) to the exact value of P(>7) found above.
 
Simple, but correct. The expectation value of a binomial distribution is p*n = (0.43)*(19) = 8.17

I will re-write the binomial probability to look better, and to put in p and n:

\(\displaystyle \displaystyle P(x) = \binom{19}{19-x}\ (0.43)^x\ (0.57)^{19-x}\)

You are trying to find P(>7). You have two choices:

either ...P(>7) = P(8) + P(9) + . . . + P(19)

or .........P(>7) = 1 - P(<7) = 1 - [P(0) + P(1) + . . . + P(7)]

The first form requires you to calculate 12 probabilities, while the second "only" needs 8. I would use the second. This calculation gives you the exact probability, but takes a lot of calculation. Often we can (or should) use a "normal approximation" to the binomial distribution.

You must have been given some empirical rules to guide when the normal approximation is good. I don't know what you were taught - look at your text or notes to see whether choice 1 (Yes) or choice 3 (No) fits the rules you have.

The standard deviation of a binomial also has a form you should memorize:

\(\displaystyle \sigma = \sqrt{n\ p\ (1-p)} = \sqrt{19(0.43)(0.57)} = 2.16 \)

Thus the normal distribution that is an approximation to this binomial has mean 8.17 and standard deviation 2.16.

That means that the z-value you use will correspond to 7.5 Hobbits, because any number greater than 7.5 will round to be >7. Find z and look in a table of the normal distribution to find the probability of >z.

Compare this result (the normal approximation) to the exact value of P(>7) found above.

I edited my initial post with what I have worked out for this problem. Can you check it over and let me know if I have it right or wrong. Thanks for the help!
 
I edited my initial post with what I have worked out for this problem. Can you check it over and let me know if I have it right or wrong. Thanks for the help!
The most probable number has to be an integer, but the "expected value" is an average which I would leave as non-integer, 8.17

Your expression for the standard deviation looks like a - sign under the radical. Should be + and the value after taking the square root is 2.16.

Rounding the probability to 0.6435 is appropriate.

Since you state that the normal approximation is appropriate, you MUST get the same answer as above!


(7.5-8.17)/2.16= 0.3101851852...OK
.31 on Z table is .1217...
?? This must be the area between 0 and -0.31
5-.1217= 4.8783...
?? Add the area from 0 to infinity: 0.5000+0.1217 = 0.6217

BUT BECAUSE THIS IS A PROBABILITY IT WOULD STILL BE .1217?
Since 7.5 is below the mean, z will be negative: z = (7.5 - 8.17)/2.16 = -0.3102
That is the number of standard deviations below the mean
My table of the normal distribution only shows the positive half, so I have to look up z=+0.31, which is F(.31) = 0.6217,
By symmetry, the area of the distribution above -0.31 is the same as the area above -0.31:
the probability P(>7), using the normal approximation, is 0.6217

Close enough?
 
Simple, but correct. The expectation value of a binomial distribution is p*n = (0.43)*(19) = 8.17

I will re-write the binomial probability to look better, and to put in p and n:

\(\displaystyle \displaystyle P(x) = \binom{19}{19-x}\ (0.43)^x\ (0.57)^{19-x}\)

You are trying to find P(>7). You have two choices:

either ...P(>7) = P(8) + P(9) + . . . + P(19)

or .........P(>7) = 1 - P(<7) = 1 - [P(0) + P(1) + . . . + P(7)]

The first form requires you to calculate 12 probabilities, while the second "only" needs 8. I would use the second. This calculation gives you the exact probability, but takes a lot of calculation. Often we can (or should) use a "normal approximation" to the binomial distribution.

You must have been given some empirical rules to guide when the normal approximation is good. I don't know what you were taught - look at your text or notes to see whether choice 1 (Yes) or choice 3 (No) fits the rules you have.

The standard deviation of a binomial also has a form you should memorize:

\(\displaystyle \sigma = \sqrt{n\ p\ (1-p)} = \sqrt{19(0.43)(0.57)} = 2.16 \)

Thus the normal distribution that is an approximation to this binomial has mean 8.17 and standard deviation 2.16.

That means that the z-value you use will correspond to 7.5 Hobbits, because any number greater than 7.5 will round to be >7. Find z and look in a table of the normal distribution to find the probability of >z.

Compare this result (the normal approximation) to the exact value of P(>7) found above.

The most probable number has to be an integer, but the "expected value" is an average which I would leave as non-integer, 8.17

Your expression for the standard deviation looks like a - sign under the radical. Should be + and the value after taking the square root is 2.16.

Rounding the probability to 0.6435 is appropriate.

Since you state that the normal approximation is appropriate, you MUST get the same answer as above!


Since 7.5 is below the mean, z will be negative: z = (7.5 - 8.17)/2.16 = -0.3102
That is the number of standard deviations below the mean
My table of the normal distribution only shows the positive half, so I have to look up z=+0.31, which is F(.31) = 0.6217,
By symmetry, the area of the distribution above -0.31 is the same as the area above -0.31:
the probability P(>7), using the normal approximation, is 0.6217

Close enough?

I had a simple addition error and recalculated the binomial probability to 0.6176.
This still isn't the same as the normal approximation probability of 0.6217. Do they have to be exactly the same?
 
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