Probability of obtaining a sample mean of x in n trials?

[moonman239]

Homework Statement

Let's say I flip a coin 80 times. I keep track of what happens when I flip it twice. What is the probability that I will find that, on the average, x heads happen per trial?

Homework Equations

The Attempt at a Solution

Determine what values x can hold in each trial in order to obtain a mean of X. Then I can use the binomial distribution on those values. Add the probabilities up. That gives me my probability.

 

[vacuum]

you are correct in your approach to solving this problem. First, we need to determine the possible values of x that can occur in each trial. In this case, since we are flipping a coin twice, the possible values are 0, 1, or 2.

Next, we can use the binomial distribution to calculate the probability of getting x heads in each trial. The binomial distribution is given by P(x) = nCx * p^x * (1-p)^(n-x), where n is the total number of trials, x is the number of successes, and p is the probability of success in each trial. In this case, n = 2, p = 0.5 (since we have a fair coin), and x can take on the values of 0, 1, or 2.

Therefore, the probability of getting 0 heads in a trial is P(0) = 2C0 * 0.5^0 * (1-0.5)^(2-0) = 0.25. The probability of getting 1 head in a trial is P(1) = 2C1 * 0.5^1 * (1-0.5)^(2-1) = 0.5. And the probability of getting 2 heads in a trial is P(2) = 2C2 * 0.5^2 * (1-0.5)^(2-2) = 0.25.

Now, to calculate the overall probability of getting an average of x heads per trial, we need to add up the probabilities of getting x heads in each trial and divide by the total number of trials (80 in this case). So the final probability would be:

P(x heads per trial) = (P(0) + P(1) + P(2)) / 80

= (0.25 + 0.5 + 0.25) / 80

= 1 / 80

= 0.0125

Therefore, the probability of getting an average of x heads per trial is 0.0125 or 1.25%.

I hope this helps and please let me know if you have any further questions. Good luck with your studies!