A certain stock's daily percent return on Fridays has a mean o...

Question

A certain stock's daily percent return on Fridays has a mean of 3.12% and a standard deviation of 41.25%. If random samples of 40 days are selected and the mean return for each sample is calculated, what is the probability that a sample mean is greater than 17%?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a certain stock's daily percent return on Fridays has a mean of 3.12% and a standard deviation of 41.25%. If random samples of 40 days are selected and the mean return for each sample is calculated, what is the probability that a sample mean is greater than 17%? And we're given four possible choices as our answers. For choice A, we have 0.0112. For choice B, we have 0.0127. For choice C, we have 0.0167, and for choice D, we have 0.1639. Now, we're asked to find the probability that a sample mean is greater than 17%. So, the first step is to convert that 17% into a Z score, and in order to do that, we'll need to have the mean of sample means, as well as the standard deviation of sample means. Now recall that according to the central limit theorem, if random samples of size N, where N is greater than or equal to 30, are drawn from any population with a mean u and a standard deviation of sigma, then the sampling distribution of mean samples approximates a normal distribution. So, if we look at what we have here, we have random samples of 40 days. That means that our in value, in our case, is going to be equal to 40, which is actually greater than 30. Which means the sampling distribution of sample means approximates a normal distribution. Therefore, our mean of sample means, we label as mu sub X bar is just gonna be equal to the population mean you. Which in this case, is going to be equal to the 3.12%, which will convert into a decimal by dividing by 100, that's gonna be equal to 0.0312. Likewise, for our standard deviation of sample means, which we will label as sigma sub X bar, that is going to be equal to sigma divided by the square root of N. or sigma here is our population standard deviation. And we're actually given that in the problem as well, that is the 41.25%, so that means that this is going to be equal to. 0.4125. Divided by the square root of 40. And when we evaluate this expression, this is going to be approximately equal to. 0.0. 652 2, and here we just rounded 25 decimal places. So, with these two values, we can now calculate our Z score for the 17%. So recall your formula for calculating a Z score, that's gonna be Z, it's going to be equal to. X bar. Minus new sub X bar, and that quantity is going to be divided by sigma sub x bar. Where export here is going to be the sample mean that we're interested in, which in this case is going to be 17%. And so we can substitute in all all of our values, so that means Z is going to be equal to. Or 17% we'll have 0.17. Once we divide by 100, convert that from a percentage into a decimal, then we'll have minus our new sub X bar, which is 0.0312, and that quantity is going to be divided by our sigma sub X bar, which is 0.06522. And when we evaluate this expression, this is going to be approximately equal to. 2.128. So here we just rounded to 3 decimal places. So next we need to find the cumulative probability associated with this Z score. And so we're gonna need to look up that value in our standard normal distribution tables. And when we look up this value in our table, we see that it falls between two of the entries. It falls between Z equals 2. 2.12. Which has a cumulative probability of. 0.9830. And the other entry is Z is equal to 2.13, which has a cumulative probability of 0.9834. So, to find the cumulative probability associated with RZ value, we'll actually have to use interpolation. So, we'll be looking for the cumulative probability E of Z less than. 2.128. And this is going to be equal to, and here we call your equation for interpolation. So, the first term here is going to be our change in our cumulative probabilities, divided by our change in Z scores from our table. So, in the numerator, we'll have the quantity of 0.9834 minus 0.9830, and that quantity is going to be divided by the quantity of 2.13 minus 2.12. In quantity. And then that fraction is going to get multiplied by the quantity of the Z score that we're looking for, which is going to be 2.128. Minus the smaller of the two Z scores from the table, so that's gonna be minus 2.12 in quantity, then we'll have to add to this the cumulative probability from the Z score for the 2.12. And so that's gonna be 0.9830. And when we evaluate this expression, this is going to be approximately equal to. 0.9833. So here we just rounded to 4 decimal places. Now, what we're actually interested in is the probability that the sample mean is greater than 17%. So we're looking for P X bar is greater than 0.17. But that is going to be equal to. P of Z greater than. 2.128. Since the score associated with the 70% was the 2.128. And so, in order to find this probability using our Z score, we just need to take the complement of the cumulative probability that we found earlier. So, this is going to be equal to 1 minus E of Z less than. 2.128. And so when we substitute in those values, that's going to be equal to 1 minus. 0.9833. Which is going to be equal to 0.0167. But this here is the probability that we're looking for for this problem, and when we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice C. So, I hope that this explanation has been useful, and I'll see you in the next video.

A certain stock's daily percent return on Fridays has a mean of 3.12% and a standard deviation of 41.25%. If random samples of 40 days are selected and the mean return for each sample is calculated, what is the probability that a sample mean is greater than 17%?

Key Concepts

Central Limit Theorem

Standard Error of the Mean

Z-Score

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