I am stuck on part c and d. I assumed part c is asking for the probability of ge
ID: 3043761 • Letter: I
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I am stuck on part c and d. I assumed part c is asking for the probability of getting a 0 in every sample, but I am not sure.
A sampling distribution (e.g. of the sample mean) is a distribution, not a histogram of observed sample means; the histogram of sample means discussed in class is just an intuitive way of thinking about the sampling distribution; technically, it'scalled the empirical* sampling distribution. Of course, if the number of trials is infinite, then the empirical sampling distribution (i.e., the histogram) approaches the distribution. Anyway, to show that the sampling distribution is truly a distibution (nt a histogram), let's derive one mathematically- no data at al 0,1, following the Consider a population described by a Bernoulli random variable, i.e.x Berngulli distribution, i .e., p(x) = p11x (1-pi)^(1-x) . Suppose we take samples of size 2 a) 1 te down all the possible samples. Hint: there are only 4 b) For each of the possible samples, compute the sample mean. C) For each of the possible samples, compute the probability. Hint: Use Bernoulli d) Based on your answens to parts a-c, find the probability of each of the possible sample means .Explanation / Answer
a) All the four possible samples are
Sample 1: (0,0)
Sample 2: (0,1)
Sample 3: (1,0)
Sample 4: (1,1).
b) Sample means are:
Sample 1: Mean=(0+0)/2=0
Sample 2: Mean=(0+1)/2=0.5
Sample 3: Mean=(1+0)/2=0.5
Sample 4: Mean=(1+1)/2=1
c) The probability of each sample is:
Sample 1: Probability = P(0)*P(0) = (1-p)*(1-p) = (1-p)2
Sample 2: Probability = P(0)*P(1) = (1-p)*p = p(1-p)
Sample 3: Probability = P(1)*P(0) = p*(1-p) = p(1-p)
Sample 4: Probability = P(1)*P(1) = p*p = p2
d) Sample mean Probability
0 (1-p)2
0.5 2p(1-p)
1 p2
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