Central Limit Theorem Question

[dirtybiscuit]

Homework Statement

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?

Homework Equations

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

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.

 

[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 , for example.

ehild

 

[STEMucator]

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

 

[mfb]

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

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.

 

[Ray Vickson]

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

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.

 

[Ray Vickson]

Homework Statement

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?

Homework Equations

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

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.

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
$$G(x) = \int_x^{\infty} \phi(t) \, dt \\<br /> A_1(x) = \frac{\phi(x)}{x} \;\; \text{approximation 1}\\<br /> A_2(x) = \frac{\phi(x)}{x} - \frac{\phi(x)}{x^3} \;\; \text{approximation 2}$$
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:

  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 $d \phi(t)/dt = - t \phi(t)$:
$$G(x) = \int_x^{\infty} \phi(t) \, dt = \int_x^{\infty} \frac{1}{t} \left(- \frac{d \phi(t)}{dt} \right) \, dt<br /> = \frac{\phi(x)}{x} - \int_x^{\infty} \frac{\phi(t)}{t^2} \, dt$$
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.