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I have a question like this.

A company manufactures light bulbs. The life time of bulbs is assumed to be normally distributed. The CEO claims that an average light bulb lasts $300$ days. A researcher randomly selected $64$ bulbs for testing and found that the mean is $290$ days and the standard deviation is $50$ days. If the CEO's claim is true, what is the probability that a randomly selected bulb would have an average life of no more that $290$ days?

I went on like this;

According to the researcher, $N(290, 50)$

But as the sample size is $64$, the population SD is $\sqrt{50^2\times64}$

So according to CEO, $N(300, 400)$

Can I really assume the population standard deviation like that using a reverse method?

Jimmy Dur
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Padmal
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  • Why have you multiplied by $64$? The researcher's result of the standard deviation being $50$ days appears to be a statement about the sample (and thus an estimate for the population) rather that the sample mean. – Henry Jan 15 '16 at 08:18
  • Yes but when we are calculating the standard deviation of a sample, we divide the population variance by the number of samples in the sample and take the square root right? So what the researcher got isn't the sample SD than the population SD since the CEO may be talking about the whole population? – Padmal Jan 15 '16 at 08:26
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    No - you are thinking about the standard error of the mean estimated from the sample rather than the standard deviation. Apart perhaps for a $\sqrt{\frac{n}{n-1}}$ adjustment to reduce bias, the sample standard deviation is a reasonable estimator of the population standard deviation – Henry Jan 15 '16 at 08:34

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Since we are interested in the probability asosciated with a single light bulb the central limit theorem is not needed here. The sample standard deviation of $50$ is already the best estimate of the population standard deviation (look up the maximum liklihood principle for more information on this). The distribution you should use is $N(300,50)$.

Spencer
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    the best estimate (max likelihood) for the variance of a gaussian distribution is $\frac{1}{N-1} \sum_{n=1}^N (x_n-\hat{\mu})^2$, so why not use $\frac{63}{64} 50^2$ as an estimation of the variance ? – reuns Jan 15 '16 at 09:11