0
$\begingroup$

Question:

Marks obtained by certain number of students are assumed to be normally distributed with mean 65 and variance 25. If three students are taken at random, what is the probability that exactly two of them will have marks over 70?

The textbook way to solve it is: Finding the probability (p) that a student gets more than 70 marks. Then find $3(C)2 * p^2 * q$

To find the probability(p) the solution first calculates z=$(70-65)/5$

My confusion is that why did it use the standard deviation of the population(5) instead of using mean of the sampling distribution of sample mean which would have been $5/sqrt(3)$?

In general how do I know when to use what because a lot of questions related to normal distribution first calculate the standard error of mean to calculate the z score.

$\endgroup$
  • $\begingroup$ What does this have to do with the sample mean? The condition is that exactly two have marks over $70$. That says nothing about the sample mean. If the question was about the sum of those students' marks, you might look at the distribution of the sample mean. $\endgroup$ – Robert Israel Nov 26 '15 at 22:25
  • $\begingroup$ stattrek.com/sampling/sampling-distribution.aspx Consider the question given in the above link in Example 1: It requires to calculate weight of average student but it uses standard error of mean to calculate the z score rather than the population s.d like in this similar question. Why? $\endgroup$ – conquester Nov 26 '15 at 22:39
0
$\begingroup$

In the stattrek example, you are not told the true distribution of the population of students. You only know the average weight and the sd. I take a sample of 50 students and then take the average of these 50 students. I call that $\bar X$. I need to know what is the probability that the average weight $\bar X$ of a sampled student will be less than 75 pounds. Well, when you take the average of 50 students, you are adding up 50 different draws from the population. The CLT implies that $\bar X$ will (approximately) follow a normal distribution. In order to use the normal approximation, you need the average of $\bar X$ and the standard distribution of $\bar X$. It turns out that the average of $\bar X$ is the true population average $$\text{avg of }\bar X = \mu_{\bar X} = 80 \text{ lbs.}$$ Using higher statistics, you can show that the standard deviation of $\bar X$ is $$\sigma_{\bar X} = \frac{20}{\sqrt{50}}.$$ I ignored the correction factor. I believe this is called the square root law. It is just a fact you have to take for granted in lower stats. Now, you have the average of $\bar X$ and the sd of $\bar X$. Next, it is just a matter of using the usual techniques to use the normal approximation.

$\endgroup$
  • $\begingroup$ No problem. Up votes and check marks would be greatly appreciated. :) $\endgroup$ – Em. Nov 26 '15 at 23:27
  • $\begingroup$ Right now I don't have the reputation to upvote but as soon as I have it, I will come back to upvote it :) $\endgroup$ – conquester Nov 26 '15 at 23:57
0
$\begingroup$

Since you were given information about all the students taking the test, you know the true population mean and standard deviation. In other words, the problem is trying to get you to relate a score of 70 to the area under the curve. The area under the curve to the left of 70 represents the percentage $p$ of students who got a 70 or worse. In other words, if I randomly pick a student from the class, the chance that they did worse than 70 is $p$. A similar case can be made for the right side of 70. Maybe in terms of a diagram, $$\text{test score} \Longleftrightarrow z\text{-score} \Longleftrightarrow \text{area under cruve } p.$$

Finally, since each student is independent of each other, you have $n=3$ independent trials with probability $p$ of success. This is a binomial distribution. Hence, the answer the book provided.

$\endgroup$
  • $\begingroup$ Thanks for answering. I have posted my doubt as a comment to my question. $\endgroup$ – conquester Nov 26 '15 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.