There are twelve green and eight yellow balls in a bag. One ball is drawn from the bag at random and

replaced. This is repeated until a yellow ball is drawn. The number of balls drawn in total is X.

(a) Find the mean and variance of X.

(b) Find the probability that X is at most one standard deviation from E(X)

I have found the mean as 5/2 and the variance as 15/4.

I’m not sure how to do part b. I know that with a normal distribution it would be 68 percent but I’m not sure how to do it with a negative binomial. Any help would appreciated. Thanks

replaced. This is repeated until a yellow ball is drawn. The number of balls drawn in total is X.

(a) Find the mean and variance of X.

(b) Find the probability that X is at most one standard deviation from E(X)

I have found the mean as 5/2 and the variance as 15/4.

I’m not sure how to do part b. I know that with a normal distribution it would be 68 percent but I’m not sure how to do it with a negative binomial. Any help would appreciated. Thanks

Original post by Cyanforest

There are twelve green and eight yellow balls in a bag. One ball is drawn from the bag at random and

replaced. This is repeated until a yellow ball is drawn. The number of balls drawn in total is X.

(a) Find the mean and variance of X.

(b) Find the probability that X is at most one standard deviation from E(X)

I have found the mean as 5/2 and the variance as 15/4.

I’m not sure how to do part b. I know that with a normal distribution it would be 68 percent but I’m not sure how to do it with a negative binomial. Any help would appreciated. Thanks

replaced. This is repeated until a yellow ball is drawn. The number of balls drawn in total is X.

(a) Find the mean and variance of X.

(b) Find the probability that X is at most one standard deviation from E(X)

I have found the mean as 5/2 and the variance as 15/4.

I’m not sure how to do part b. I know that with a normal distribution it would be 68 percent but I’m not sure how to do it with a negative binomial. Any help would appreciated. Thanks

Its really a geometric distribution (just one win), but you have the mean and standard deviation so you can look at the probabilities which correspond to values

[mean - std dev, mean + std dev]

There are only ~4 values so you could compute the probabilities and sum them up or use the cumulative geometric

https://en.wikipedia.org/wiki/Geometric_distribution

(edited 8 months ago)

Original post by mqb2766

Its really a geometric distribution (just one win), but you have the mean and standard deviation so you can look at the probabilities which correspond to values

[mean - std dev, mean + std dev]

There are only ~4 values so you could compute the probabilities and sum them up or use the cumulative geometric

[url=https://en.wikipedia.org/wiki/Geometric_distribution]https://en.wikipedia.org/wiki/Geometric_distribution]https://en.wikipedia.org/wiki/Geometric_distribution

[mean - std dev, mean + std dev]

There are only ~4 values so you could compute the probabilities and sum them up or use the cumulative geometric

[url=https://en.wikipedia.org/wiki/Geometric_distribution]https://en.wikipedia.org/wiki/Geometric_distribution]https://en.wikipedia.org/wiki/Geometric_distribution

I have tried finding the probability that x is between 8 and 12 inclusive which gave me 0.0258 (3sf) I’m not sure what to do next to relate this to the standard deviation from the mean? (The answer is 0.8704)

Original post by Cyanforest

I have tried finding the probability that x is between 8 and 12 inclusive which gave me 0.0258 (3sf) I’m not sure what to do next to relate this to the standard deviation from the mean? (The answer is 0.8704)

As thats the main body of the distribution (mean+/-sigma), 0.02 is way too small. What did you actually do? The mean is 2.5 and the stddev ~1.9 so not sure where 8 and 12 comes from?

Edit - also note that the ans comes out of the cumulative directly, without having to sum the individual probabilities.

(edited 8 months ago)

Original post by mqb2766

As thats the main body of the distribution (mean+/-sigma), 0.02 is way too small. What did you actually do? The mean is 2.5 and the stddev ~1.9 so not sure where 8 and 12 comes from?

Ah okay I took the 12 and 8 from the amount of green and yellow balls but that makes no sense. I think I’m confused about what I’m actually meant to do regarding the mean and standard deviation and using the cumulative geometric distribution. 2.5 +/- 1.9 = 4.4/0.6 and I’m not sure how that relates to the probability.

Original post by Cyanforest

Ah okay I took the 12 and 8 from the amount of green and yellow balls but that makes no sense. I think I’m confused about what I’m actually meant to do regarding the mean and standard deviation and using the cumulative geometric distribution. 2.5 +/- 1.9 = 4.4/0.6 and I’m not sure how that relates to the probability.

Yup, thats simply wrong.

E(X) is the mean of the distribution so roughly the "centre". The std dev is the "width" (or half the width of our interval). So were interested in

X in [mean - std dev, mean + std dev]

obv X is discrete but that the interval is roughly [0.5, 4.5] so we want the discrete values in that and either compute the relevant probs and sum them, or use the relevant cumulative.

The mean and std dev give you an idea about where the main body of the distribution is and its width, irrespective of the distribution. Thats a key thing to understand. Tbh, Id get in the habit of sketching the distribution (even if the question doesnt ask for it) so that you force yourself to be clear about what youre actually analysing.

(edited 8 months ago)

Original post by mqb2766

Yup, thats simply wrong.

E(X) is the mean of the distribution so roughly the "centre". The std dev is the "width" (or half the width of our interval). So were interested in

X in [mean - std dev, mean + std dev]

obv X is discrete but that the interval is roughly [0.5, 4.5] so we want the discrete values in that and either compute the relevant probs and sum them, or use the relevant cumulative.

The mean and std dev give you an idea about where the main body of the distribution is and its width, irrespective of the distribution. Thats a key thing to understand. Tbh, Id get in the habit of sketching the distribution (even if the question doesnt ask for it) so that you force yourself to be clear about what youre actually analysing.

E(X) is the mean of the distribution so roughly the "centre". The std dev is the "width" (or half the width of our interval). So were interested in

X in [mean - std dev, mean + std dev]

obv X is discrete but that the interval is roughly [0.5, 4.5] so we want the discrete values in that and either compute the relevant probs and sum them, or use the relevant cumulative.

The mean and std dev give you an idea about where the main body of the distribution is and its width, irrespective of the distribution. Thats a key thing to understand. Tbh, Id get in the habit of sketching the distribution (even if the question doesnt ask for it) so that you force yourself to be clear about what youre actually analysing.

Thank you I understand it now

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