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i dont understand how they did part e

Why did they do P(M=0)= 27/125+ 54/125

and why did they do P(M=1)

i dont even understand the question as well what does the sampling distribution of mode M even mean.

Why did they do P(M=0)= 27/125+ 54/125

and why did they do P(M=1)

i dont even understand the question as well what does the sampling distribution of mode M even mean.

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here is the question and worked example

**assassinbunny123**)here is the question and worked example

**sampling distribution**is a probability

**distribution**of a statistic obtained through a large number of samples drawn from a specific population. The

**sampling distribution**of a given population is the

**distribution**of frequencies of a range of different outcomes that could possibly occur for a statistic of a population.

So, in this case - the mode is the statistic. Every sample from the population is different and therefore will give a different mode. All the different modes will form a range of modes that the population can take. The probability associated with each mode from each sample forms the probability distribution of the mode. This is known as the sampling distribution of the mode.

Clearly, there are two possibilities of numbers that the counters can have: 0 or 1.

A simple random sample of size 3 is taken from this population.

E.g. (0,1,1). It can also be (0,0,0), (1,1,1), (1,0,1) etc.

So, as you can see - different samples from the same population give a different mode.

(0,1,1) --> mode = 1.

(0,0,0) --> mode = 0, etc.

In part (b), you listed all the possible samples that can possibly be taken from the population.

Work out the modes for each one and the associated probability of getting that sample e.g. (0,0,0) = 0.6 * 0.6 * 0.6 = (0.6^3) because there is a probability of 0.6 of getting a 0.

The example does it very nice in part (c), they listed the cases next to the probabilities. It is clear that the first two lines, i.e. (0,0,0) case and the next line all have a mode of 0.

If you add the probabilities of those samples, you get 81/125.

The other possibility is a mode of 1. The probabilities always add up to 1. So P(M = 1) = 1 - (81/125).

On the other hand, you could use the last two lines of part (c) to find the probability of getting a mode of 1. Then, subtract this from 1 to find the probability of getting a mode of 0.

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If i ever meet you in real life i'll give you a hug. thank you very much for this valuable information

(Original post by

A

So, in this case - the mode is the statistic. Every sample from the population is different and therefore will give a different mode. All the different modes will form a range of modes that the population can take. The probability associated with each mode from each sample forms the probability distribution of the mode. This is known as the sampling distribution of the mode.

Clearly, there are two possibilities of numbers that the counters can have: 0 or 1.

A simple random sample of size 3 is taken from this population.

E.g. (0,1,1). It can also be (0,0,0), (1,1,1), (1,0,1) etc.

So, as you can see - different samples from the same population give a different mode.

(0,1,1) --> mode = 1.

(0,0,0) --> mode = 0, etc.

In part (b), you listed all the possible samples that can possibly be taken from the population.

Work out the modes for each one and the associated probability of getting that sample e.g. (0,0,0) = 0.6 * 0.6 * 0.6 = (0.6^3) because there is a probability of 0.6 of getting a 0.

The example does it very nice in part (c), they listed the cases next to the probabilities. It is clear that the first two lines, i.e. (0,0,0) case and the next line all have a mode of 0.

If you add the probabilities of those samples, you get 81/125.

The other possibility is a mode of 1. The probabilities always add up to 1. So P(M = 1) = 1 - (81/125).

On the other hand, you could use the last two lines of part (c) to find the probability of getting a mode of 1. Then, subtract this from 1 to find the probability of getting a mode of 0.

**Chittesh14**)A

**sampling distribution**is a probability**distribution**of a statistic obtained through a large number of samples drawn from a specific population. The**sampling distribution**of a given population is the**distribution**of frequencies of a range of different outcomes that could possibly occur for a statistic of a population.So, in this case - the mode is the statistic. Every sample from the population is different and therefore will give a different mode. All the different modes will form a range of modes that the population can take. The probability associated with each mode from each sample forms the probability distribution of the mode. This is known as the sampling distribution of the mode.

Clearly, there are two possibilities of numbers that the counters can have: 0 or 1.

A simple random sample of size 3 is taken from this population.

E.g. (0,1,1). It can also be (0,0,0), (1,1,1), (1,0,1) etc.

So, as you can see - different samples from the same population give a different mode.

(0,1,1) --> mode = 1.

(0,0,0) --> mode = 0, etc.

In part (b), you listed all the possible samples that can possibly be taken from the population.

Work out the modes for each one and the associated probability of getting that sample e.g. (0,0,0) = 0.6 * 0.6 * 0.6 = (0.6^3) because there is a probability of 0.6 of getting a 0.

The example does it very nice in part (c), they listed the cases next to the probabilities. It is clear that the first two lines, i.e. (0,0,0) case and the next line all have a mode of 0.

If you add the probabilities of those samples, you get 81/125.

The other possibility is a mode of 1. The probabilities always add up to 1. So P(M = 1) = 1 - (81/125).

On the other hand, you could use the last two lines of part (c) to find the probability of getting a mode of 1. Then, subtract this from 1 to find the probability of getting a mode of 0.

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(Original post by

If i ever meet you in real life i'll give you a hug. thank you very much for this valuable information

**assassinbunny123**)If i ever meet you in real life i'll give you a hug. thank you very much for this valuable information

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