# Thread: Statistics Problem - Normal Dist. - Mean & Stdev Given

1. ## Statistics Problem - Normal Dist. - Mean & Stdev Given

Hey all,

I'm new to this forum and I have been struggling with this questiong for a while:

Imagine you are building a wall from precast concrete blocks. Standard 20 cm Blocks are 19.5 cm high to allow for a 0.5 cm layer of mortar. In practice, the height of a block plus mortar row varies according to a Normal Distribution with mean 20.0 cm and standard deviation 0.25 cm. Heights of successive rows are independent. Your wall has 10 rows.
(i) What is the distribution of the height of the wall? ( 3 marks)
(ii) What is the probability that the height of the wall differs from the design height of 200 cm by more than 1 cm? ( 4 marks)

I'm just really confused about how to approach it.

Because it is a sample of 10 from a normal distribution, can you apply the t distribution with sample mean 20.0cm and sample stdev. 0.25cm?

From this, I'm not quite sure how to approach part (ii)

Any guidance would be greatly appreciated.

2. The sum of normally distributed variables is also normal.

Work out the mean and standard deviation of $y = (x_1 + x_2 +x_3 +...x_{10})$ and use those as the paramters for the distribution of the wall.

Hint: for independant variables $x_1,x_2$
$Var(x_1+x_2) = var(x_1) + var(x_2)$

part (ii)
Once you have found the distribution of y
You want P(y>204) + P(y<196)

If you dont remember how to do that, you need to standardise the normal variable.

Ill do the first one:
$y \sim N(a,b^2)$
Define: $Z=\frac{y-a}{b}$
$Z \sim N(1,0)$

Now:
$p(y \geq 204) = 1 - p(y \leq 204)$
$=1 - p \left(\frac{y-b}{a} \leq \frac{204-a}{b} \right)$

$p(y \geq 204) = 1 - p \left(Z \leq \frac{204-a}{b} \right)$
The expression in brackets is jus a number, you can look it up in your standard normal table.

You should be able to do p(y<196) from that method or by general reasoning. If you are totally stuck you can read the spoiler below

Spoiler:

The mean of y is 200, and the normal distribution is symetric about the mean.

This means that p(y>204) = p(y<196)

So because I am not given the values for the 10 samples, the mean of these will be the same as the mean for the whole normal distribution?

And to calculate the standard deviation should I find the sqrt of the variance of the sample of 10 values (which is 10*(Stdev of Normal Dist.)^2) ?

Regards,

Matt

4. the mean of each sample value is 20. (i think thats what you meant). The mean height of the wall is 20*10 = 200

Your method for the std.dev is correct too

5. I have no problem using Z scores, however, I do not understand why you have chosen to use P(y>204) + P(y<196) when the question asks what is the probability of the height of the wall differs from the design height by more than 1cm?

I would have used P(y<199) and P(y>201).

Am I blatantly missing something?

Thanks,

Matt

6. Also, my answer to (i) is Sample Dist ~ N (200, 0.79^2)

7. yes i think i misread the question on part b

8. Thanks mate.