Homework Help: Central Limit Theorem Question

1. May 4, 2014

dirtybiscuit

1. The problem statement, all variables and given/known data
The Rockwell hardness of certain metal pins is known to have a mean of 50 and a standard deviation of 1.5.
a)if the distribution of all such pin hardness measurements is known to be normal, what is the probability that the average hardness for a random sample of 9 pins is at least 52.
b)what if sample is 40 pins?

2. Relevant equations
So I know that μ = 50 and σ(original) = 1.5, and n = 9

3. The attempt at a solution
σ(sample) = σ(original)/$\sqrt{9}$ = 1.5/3 = .5

now find P($\bar{x}$ > 52) by relating it to the standard normal dist

find z = (52 - 50)/.5 = 4.

This seems like a huge z score to have and it pretty much much makes the probability 0. Am I doing something wrong here? It just seems like a pointless question if in both cases the probability comes to zero.

Last edited: May 4, 2014
2. May 4, 2014

ehild

The probabilities are really very small, but still differ from zero. There are online calculators http://www.solvemymath.com/online_math_calculator/statistics/cdf_calculator.php [Broken], for example.

ehild

Last edited by a moderator: May 6, 2017
3. May 4, 2014

Zondrina

Your sample size is quite small. Shouldn't you be using a t-estimator instead?

4. May 4, 2014

Staff: Mentor

We know (or assume) that all pins follow a normal distribution. Then the mean of 9 pins follows a normal distribution as well.
The calculation is fine, the result is just a small number.

5. May 4, 2014

Ray Vickson

No. When the variance is known we use the normal distribution. When the variance is unknown (and is estimated using the data itself) we use the t-distribution.

6. May 4, 2014

Ray Vickson

The probability is small, but nowhere near zero. (In fact, the ability to estimate very small probabilities is crucial in reliability and safety studies, etc.) For large $x > 0$ you can find simple, fairly accurate approximations to $G(x) \equiv P(X > x)$ for standard normal $X$.
Let $$\phi(t) = \frac{1}{\sqrt{2 \pi}} e^{-t^2/2}.$$
We have
$$G(x) = \int_x^{\infty} \phi(t) \, dt \\ A_1(x) = \frac{\phi(x)}{x} \;\; \text{approximation 1}\\ A_2(x) = \frac{\phi(x)}{x} - \frac{\phi(x)}{x^3} \;\; \text{approximation 2}$$
The functions $A_1(x)$ and $A_2(x)$ are approximations to $G(x)$, with $A_2$ being a bit better than $A_1$ for large $x > 0$ (but may be worse for small $x > 0$). You can see how they perform from the following table:
Code (Text):

x       G(x)      A1(x)      A2(x)
1.0  1.58655e-01 2.41971e-01 0.00000e+00
1.5  6.68072e-02 8.63451e-02 4.79695e-02
2.0  2.27501e-02 2.69955e-02 2.02466e-02
2.5  6.20967e-03 7.01132e-03 5.88951e-03
3.0  1.34990e-03 1.47728e-03 1.31314e-03
3.5  2.32629e-04 2.49338e-04 2.28984e-04
4.0  3.16712e-05 3.34576e-05 3.13665e-05

These approximations are based in integration by parts, using the fact that $d \phi(t)/dt = - t \phi(t)$:
$$G(x) = \int_x^{\infty} \phi(t) \, dt = \int_x^{\infty} \frac{1}{t} \left(- \frac{d \phi(t)}{dt} \right) \, dt = \frac{\phi(x)}{x} - \int_x^{\infty} \frac{\phi(t)}{t^2} \, dt$$
The approximation $A_1(x)$ is just the first term above, and it should be a pretty good approximation for large $x > 0$ because the magnitude of the neglected term is less than $G(x)/x^2$. The approximation $A_2(x)$ estimates the second term above through another integration by parts.

Last edited: May 4, 2014