# 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