How Do You Calculate the Probability of a Sample Mean in Normal Distribution?

AI Thread Summary
The discussion focuses on calculating the probability of a sample mean in a normal distribution scenario involving examination scores. The scores are normally distributed with specific means and standard deviations, leading to a final mark calculation. A participant attempts to find the probability that a sample of eight candidates has a mean final mark of less than 60 but encounters issues related to the assumption of independence between test scores. Additionally, there is a conversation about using the Poisson distribution for a different problem, with participants debating the appropriateness of approximating it with a normal distribution due to the Central Limit Theorem. The importance of using the cumulative distribution function (CDF) for the Poisson distribution is emphasized as a more suitable approach for calculating probabilities.
Unusualskill
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The scores X1 and X2 in papers 1 and 2 of an examination are normally distributed with means 24.3 and 31.2 respectively and standard deviations 3.5 and 3.1 respectively The final mark for each candidate is found by calculating 2X1+1.5X2. Find the probability that a random sample of 8candiates will have a mean final mark of less than 60.

what i have done so far is F~N(95.4,70.6225) and Sample mean of F, N (95.4,70.6225/8)
P(sample mean of F<60)
... then unable to do alrdy because the z-score is too large
 
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It seems to me that independence would be violated rather badly. :-) Students scoring highly on one test are more likely to score highly on the other. You're probably supposed to assume independence, but you should scold your teacher for not being explicit about it.
 
chogg said:
It seems to me that independence would be violated rather badly. :-) Students scoring highly on one test are more likely to score highly on the other. You're probably supposed to assume independence, but you should scold your teacher for not being explicit about it.

Chogg, can i ask you another question, which my answer is different from answer given.
The random variable X is distributed Po(4.1) .Find the probability that the mean of a random sample of 50 observations of X is less than 4.5.

first, i find out p=4.1/50 variance=npq=3.7638
then approx to normal, N(4.1,3.7638/50)
P(sample mean<4.5)
continuity correction with -1/100
anything wrong?what is the ans u get?
 
I assume "Po" means Poisson, and 4.1 is the value of \lambda.

Why would you switch to the approximating normal? It's going to be a rather lousy approximation with such a small mean. Why not just use the poisson directly?

Anyway, you might find it useful to know that the sum of (independent!) draws from Poisson distributions is Poisson-distributed, and its mean is the sum of the means. You also might find it useful to translate the statement about the mean to a statement about the sum of your 50 random draws.
 
But how to do Possion with continuous(4.5) ? r u able to get z-score for this ques 1.34?if yes, can you show me the step pls thanks alot...if like u said using sum, then Po~(205) still need to use normal is it?
 
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Unusualskill said:
But how to do Possion with continuous(4.5) ?

http://en.wikipedia.org/wiki/Poisson_distribution

The output of a Poisson is discrete: 0, 1, 2, etc.

The mean of a Poisson is a continuous quantity. It could be anything.

For example, imagine a probability distribution which gives 1 with probability p, and 0 with probability 1-p. It only ever outputs discrete numbers (1 or 0), but its mean is a continuous quantity (p).

Unusualskill said:
r u able to get z-score for this ques 1.34?if yes, can you show me the step pls thanks alot...if like u said using sum, then Po~(205) still need to use normal is it?

Why do you care about z-scores? This is a Poisson problem, not a Gaussian problem. z-scores are for Gaussian distributions.

Also: what does "ques 1.34" mean? Is this a homework problem? (It sure sounds like it.) If it is, you're asking in the wrong forum.

In any case: if you want to know the probability that a random variable is less than some value, the concept that will help you is the cumulative distribution function, or "CDF". What is the CDF for the Poisson distribution? Does it give a reasonable-sounding answer?
 
chogg said:
http://en.wikipedia.org/wiki/Poisson_distribution

The output of a Poisson is discrete: 0, 1, 2, etc.

The mean of a Poisson is a continuous quantity. It could be anything.

For example, imagine a probability distribution which gives 1 with probability p, and 0 with probability 1-p. It only ever outputs discrete numbers (1 or 0), but its mean is a continuous quantity (p).



Why do you care about z-scores? This is a Poisson problem, not a Gaussian problem. z-scores are for Gaussian distributions.

Also: what does "ques 1.34" mean? Is this a homework problem? (It sure sounds like it.) If it is, you're asking in the wrong forum.

In any case: if you want to know the probability that a random variable is less than some value, the concept that will help you is the cumulative distribution function, or "CDF". What is the CDF for the Poisson distribution? Does it give a reasonable-sounding answer?

im using normal because of central limit theorem, which states that if n is large, the distribution can be approx to normal. Because i duno homework to do poisson for P(X<4.5)...
 
Unusualskill said:
im using normal because of central limit theorem, which states that if n is large, the distribution can be approx to normal. Because i duno homework to do poisson for P(X<4.5)...

Did you find the CDF for the Poisson distribution from the wikipedia page I linked to?

Did you try plugging in your numbers to that formula?
 
chogg said:
Did you find the CDF for the Poisson distribution from the wikipedia page I linked to?

Did you try plugging in your numbers to that formula?

i know that formula for poisson . P(x<4.5)=P(x=0,1,2,3,4)? but this is a question for sampling? or can you show me how you did?
 
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