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I'm currently preparing for an exam, and I just can't seem to grasp how to go about the question given the following information:

Information: A brick factory produce bricks whose length $X_i$ can be assumed normally distributed with mean length of $228 mm$ and standard deviation $4 mm$ ..

Question: At the brick factory a random sample of $50$ bricks is taken and the average of the lengths calculated by $$\bar X=\frac{1}{50}\sum^n_{i=1}X_i$$ What is the probability that the average of the observed lengths $\bar X$, is within the range of $[227;229]$.

My thoughts: In the question before this, I was asked to calculate the 0.95 CI and found it to be in the range: $220 mm < X < 236 mm$, so I would assume when we averaged the random sample of 50 bricks, the probability would be over $90$ pct.

I hope someone can point me in the right direction.

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  • $\begingroup$ Do you know what the mean and standard deviation of $\bar{X}$ are? $\endgroup$ – Dilip Sarwate Dec 12 '15 at 15:03
  • $\begingroup$ I don't know anything besides what I stated under "Information:" .. So to sum up I know the overall mean of 228 and the standard deviation 4. But I of course know nothing about the values of the random sample of 50 bricks :) $\endgroup$ – Akudo Dec 12 '15 at 15:07
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    $\begingroup$ Then you really need to read your textbook instead of trying to solve this problem. The question I asked was not about the individual values of the random sample, but about the average of the 50 values, that $\bar{X}$ in your "question" Do you know what $\bar{X}$ might be expected to be? 228 mm? 220 mm? $\endgroup$ – Dilip Sarwate Dec 12 '15 at 15:13
  • $\begingroup$ Alright, I will do that instead.. $\bar X$ is expected to be roughly 228 mm, roughly because we are estimating a proportion (50) of the original sample. $\endgroup$ – Akudo Dec 12 '15 at 15:19
  • $\begingroup$ Think about what happens when you sum up normally distributed variables? $\endgroup$ – abaumann Dec 12 '15 at 15:40
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You need to use some basic results about means and variances

  • The expectation of a sum is the sum of the expectations

  • The expectation of a constant times a random variable is the constant times the expectation of the variable

  • The variance of a sum of independent random variables is the sum of the variances

  • The variance of a constant times a random variable is the square of the constant times the variance of the variable

The two results about expectation mean that the distribution of a sample mean of identically distributed quantities has the same mean as the original variable. The two results about variance mean that the variance of a sample mean (of independent identically distributed variates) has variance that's the original variance divided by the sample size.

Hence the sample mean of a sample of 50 bricks has mean $228$ mm and standard deviation $\frac{4}{\sqrt{50}}$ mm.

  • You also need that linear combinations of independent normal random variates will also be normally distributed.

As a result, the sample mean will be $\sim N(228,0.5657)$

From there you should be able to find the probability called for in the question.

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