Highschool (year 12) homework question (statistical analysis)

probabilitystatistics

I am getting stuck on part b of the following question

A company manufactures bricks with normally distributed weights. The mean weight is 0.96kg with standard deviation 0.045kg

a) Let Y be the mean of weight of 9 such bricks. Find the distribution of Y
b) What is the probability that a random sample of 9 bricks has mean weight more than 1kg?

So my answer for a is
$Y \sim N(0.96, 0.045^{2})$

However, I don't know where to start for b). Could someone please help me with this? Thank you in advance

Best Answer

I will show you the general method for such a problem and let you plug in the specific numbers for your particular problem.

Suppose bricks have weights distributed as $\mathsf{Norm}(\mu, \sigma).$ [Notice that in this notation the second parameter is the standard deviation, not the variance.]

(a) The mean $Y = \bar X$ of $n = 9$ such bricks is distributed as $\bar X \sim \mathsf{Norm}(\mu, \sigma/\sqrt{n}).$ [That is, $SD(\bar X) = \sigma/\sqrt{n}.]$

(b) You seek $$P(\bar X > a) = P\left(\frac{\bar X - \mu}{\sigma/\sqrt{n}} > \frac{a - \mu}{\sigma/\sqrt{n}}\right) = P\left(Z > \frac{a - \mu}{\sigma/\sqrt{n}}\right),$$ where $Z$ has a standard normal distribution.

In the final expression, let $a = 1, \mu=0.96, \sigma = 0.045, n = 9.$ and use a printed table of the standard normal cumulative distribution function to get your answer.

Using R, in which $\mathtt{pnorm}$ is a normal CDF with specified parameters, one can get the numerical answer $P(\bar X > 1) = 0.00383$ as shown below. (Because of rounding, your result using printed normal tables will be slightly less precise.)

n = 9; mu = 0.96; sg = 0.045
1 - pnorm(1, mu, sg/sqrt(n))  # last param is SD
[1] 0.003830381

In the following picture, the dotted density curve is for the population of bricks, the solid curve is for the mean of $n = 1$ bricks, and the (small) area under the solid curve to the right of the vertical line represents the desired probability.

enter image description here

Notes: (1) Roughly speaking the PDF of the population is three times as wide as the PDF for the PDF of the mean of nine. Thus in order to include total probability 1 under the curve, the PDF of the mean must be three times as tall.

(2) R is excellent statistical software, available free of charge for Windows, Mac and Unix computers from here. It is easy to learn to use--provided you use just what you need at each step.

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