# Thread: Data Analysis -- Proportion Problem and Probability

1. ## Data Analysis -- Proportion Problem and Probability

According to a marketing research firm, 52% of all residential telephone numbers in Los Angeles are unlisted. A telephone sales firm uses random digit dialing equipment that dials residential numbers at random, whether or not they are listed in the telephone directory. The firm calls 500 numbers in Los Angeles.

What is the probability that at least half of the numbers dialed are unlisted?

This is one of the questions that was on a bonus quiz I took. I got it wrong and I've been trying and trying to figure it out and I just can't. I'm using a calculator, the normalcdf function so I need to know the lowerbound, upperbound, mean, and standard deviation but every time I put in what I think is the correct information, I get 0.

Here's what I was inputting:

lowerbound : 250
upperbound : 10000
mean : .52
standard deviation : .0223427841

I went over a problem the Professor did earlier and did everything she did step by step so I'm not sure if one of my figures is wrong or if I'm using the wrong function. Maybe I should be using binomialcdf? I just tried it and I'm still not getting the right answer. The only possible answers are :

.5166
.4999
.5080
.8146

I guessed the first one after trying for so long that I was nearly out of time and it was wrong so we're down to .4999, .5080, or .8146.

According to a marketing research firm, 52% of all residential telephone numbers in Los Angeles are unlisted. A telephone sales firm uses random digit dialing equipment that dials residential numbers at random, whether or not they are listed in the telephone directory. The firm calls 500 numbers in Los Angeles.

What is the probability that at least half of the numbers dialed are unlisted?

This is one of the questions that was on a bonus quiz I took. I got it wrong and I've been trying and trying to figure it out and I just can't. I'm using a calculator, the normalcdf function so I need to know the lowerbound, upperbound, mean, and standard deviation but every time I put in what I think is the correct information, I get 0.

Here's what I was inputting:

lowerbound : 250
upperbound : 10000
mean : .52
standard deviation : .0223427841

I went over a problem the Professor did earlier and did everything she did step by step so I'm not sure if one of my figures is wrong or if I'm using the wrong function. Maybe I should be using binomialcdf? I just tried it and I'm still not getting the right answer. The only possible answers are :

.5166
.4999
.5080
.8146

I guessed the first one after trying for so long that I was nearly out of time and it was wrong so we're down to .4999, .5080, or .8146.
Let X be the random variable number of unlisted numbers that are called.

X ~ Binomial(n = 500, p = 0.52)

Calculate $\Pr(X \geq 250)$.

Depending on how accurate your answer is meant to be, you might be expected to use a normal approximation to the binomial distribution to get the answer.

I get 0.8264 (correct to 4 decimal places).

3. That's odd because I think I might have gotten the same answer when I tried to do a binomialcdf but that answer isn't listed. Did I do the normalcdf incorrectly by any chance?

4. Originally Posted by mr fantastic
Let X be the random variable number of unlisted numbers that are called.

X ~ Binomial(n = 500, p = 0.52)

Calculate $\Pr(X \geq 250)$.

Depending on how accurate your answer is meant to be, you might be expected to use a normal approximation to the binomial distribution to get the answer.

I get 0.8264 (correct to 4 decimal places).
The normal approx gives p=0.814675 without a continuity correction and 0.826391 with a continuity correction.

CB

5. I'm still getting answers like 0 and 1 and I'm not sure why. It wants to know the probability that at least half of the numbers dialed are unlisted...half of 500 is 250 so that's my lowerbound, yes?

The upperbound would be 10,000, my Professor has said that you always take it to the extreme so I can't see that being wrong either.

And in the problem she did with us, the probability was equal to the mean so the mean would be .52 right?

And since this is a proportion problem, you'd have to find the square root of pi(1 - pi)/n yes? Doing that got me the aforementioned figure. And yet I still get a 0 or a 1. >.<

6. Originally Posted by mr fantastic
[snip]
I get 0.8264 (correct to 4 decimal places).
I should have mentioned that this is the answer I get using the binomial distribution.

I'm still getting answers like 0 and 1 and I'm not sure why. It wants to know the probability that at least half of the numbers dialed are unlisted...half of 500 is 250 so that's my lowerbound, yes?

The upperbound would be 10,000, my Professor has said that you always take it to the extreme so I can't see that being wrong either.

And in the problem she did with us, the probability was equal to the mean so the mean would be .52 right?

And since this is a proportion problem, you'd have to find the square root of pi(1 - pi)/n yes? Doing that got me the aforementioned figure. And yet I still get a 0 or a 1. >.<
Your sample size is 500. How can the upper bound be 10,000?

7. I don't know, she always just told us to go to 10,000. She said that even if we know, reasonably, that our upperbound cannot be 10,000, we're still supposed to go up that high. Like if we're doing test scores. We know that it can only go as high as a 100 but she still indicated that she wanted it to go to 10,000 anyways.

If I change it so that it's 500, I still get 0.

I don't know, she always just told us to go to 10,000. She said that even if we know, reasonably, that our upperbound cannot be 10,000, we're still supposed to go up that high. Like if we're doing test scores. We know that it can only go as high as a 100 but she still indicated that she wanted it to go to 10,000 anyways.

If I change it so that it's 500, I still get 0.
You will need to post all the details of your calculation if you want your mistake(s) to be pointed out.

9. Okay, well the lowerbound and upperbound have already been covered. I got the mean from a problem the Prof did earlier in which she stated that the probability is the mean so that's why I used .52. As for the standard deviation, I did the square root of:

.52(1-.52)/500

...which got me .0223427841. To calculate the probability, I did normalcdf(250, 10000, .52, .0223427841) the first time and then normalcdf(250, 500, .52, .0223427841) the second time and neither were correct.

Okay, well the lowerbound and upperbound have already been covered. I got the mean from a problem the Prof did earlier in which she stated that the probability is the mean so that's why I used .52. As for the standard deviation, I did the square root of:

.52(1-.52)/500

...which got me .0223427841. To calculate the probability, I did normalcdf(250, 10000, .52, .0223427841) the first time and then normalcdf(250, 500, .52, .0223427841) the second time and neither were correct.
You're using the wrong mean and the wrong standard deviation. For a binomial distribution the mean is np and the variance is np(1-p). As always, the standard deviation is the square root of the variance.

Stop copying your professor's examples and start understanding the work.

According to a marketing research firm, 52% of all residential telephone numbers in Los Angeles are unlisted. A telephone sales firm uses random digit dialing equipment that dials residential numbers at random, whether or not they are listed in the telephone directory. The firm calls 500 numbers in Los Angeles.

What is the probability that at least half of the numbers dialed are unlisted?

This is one of the questions that was on a bonus quiz I took. I got it wrong and I've been trying and trying to figure it out and I just can't. I'm using a calculator, the normalcdf function so I need to know the lowerbound, upperbound, mean, and standard deviation but every time I put in what I think is the correct information, I get 0.

Here's what I was inputting:

lowerbound : 250
upperbound : 10000
mean : .52
standard deviation : .0223427841
The mean is the mean number of unlisted number in a sample of 500 numbers, that is 260.

The standard deviation is sqrt(500*0,.52*0.48) which is the standard deviation of the number of unlisteds dialed.

The 250 is OK (249.5 would give the continuity corrected answer), the upper limit should be infinity and 10000 will do for that (the normal approximation extends all the way to infinity but anything beyond about mean+5*sigma is adequate for numerical calculation).

CB

12. I get 1-.1736=.8264 where I used the correction factor, i.e., $P(X>249.5)$.