# Adjusting mean and standard deviation

There's a set of 8 bags with the following weights in grams given:

1013, 997, 1013, 1013, 1004, 985, 991, 997

The mean is 1001.625, unbiased standard deviation is 10.86.

I have the following question on an exam paper:

Assuming the standard deviation is 10 and the bags have a mean weight of 1kg (1000g),
calculate the probability that five randomly selected bags together weigh more than
4.99kg.


Could someone please help me calculate this? How do I get from the weight of the bags, mean and standard deviation to the probability that five random ones weight more than something?

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Five selected from the eight? With or without replacement? Or five selected from a normally distributed population with the given mean and standard deviation? – Henry Sep 3 '12 at 13:01
What is the unbiased estimate of standard deviation? Maybe you mean the square root of the unbiased estimate of variance? – Michael Chernick Sep 3 '12 at 13:13

You need to know the distribution the bags are sampled from. If it is assumed to be normal with mean 1000 g and standard deviation 10 g then the sum of 5000 g and variance 500 g or standard deviation 10 √5. Then Z= (X-5000)/(10 √5) has a standard normal and the probability

P[X>a]=P[Z>(a-5000)/(10 √5)] which can be obtained from the table of the standard normal distribution.

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This question assumes that the weight of each bag independently follows a normal distribution with mean $\mu=1000g$ and standard deviation $\sigma=10g$. Let's refer to the weight of these five randomly chosen bags as $X_i$, $i=1,\ldots,5$. So what is the distribution of the combined weight of the five bags? That is, what distribution does $\sum_{i=1}^{5}X_i$ follow? It is well known that sums of normal random variables follow a normal distribution. The mean of this distribution will be $\mu_{\textrm{sum}}=\textrm{E}\left[\sum_{i=1}^{5}X_i\right]=\sum_{i=1}^{5}\textrm{E}\left[X_i\right]=5\mu=5000g$ and the standard deviation of the distribution will be $\sigma_{\textrm{sum}}=\sqrt{\textrm{Var}\left[\sum_{i=1}^{5}X_i\right]}=\sqrt{\sum_{i=1}^{5}\textrm{Var}\left[X_i\right]}=\sqrt{5\sigma^2}=10\sqrt{5}g$. So now what is the probability that a variable $Y\sim N\left(\mu_{\textrm{sum}}=5000g,\sigma_{\textrm{sum}}=10\sqrt{5}g\right)$ exceeds $4990g$?

\begin{align} \textrm{P}\left(Y\geq4990g\right) &= \textrm{P}\left(\frac{Y-\mu_{\textrm{sum}}}{\sigma_{\textrm{sum}}}\geq\frac{4990g-5000g}{10\sqrt{5}g}\right)\\ &= \textrm{P}\left(Z\geq-\frac{1}{\sqrt{5}}\right)\\ &= 0.6726 \end{align}

where $Z\sim N\left(0,1\right)$ is the standard normal distribution.

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