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Normal Distribution


Let us consider the following exercise: An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch.   The specification limits under which the ball bearing can operate are 0.74 inch (lower) and 0.76 inch (upper).   Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed with a mean of 0.753 inch and a standard deviation of 0.004 inch.
So we have that the diameter of the ball bearings is approximately normal with $\mu = .753$ in. and $\sigma = .004$ in.
For this problem, note that "Target" = .75, and "Actual mean" = .753.

This page will contain the following solutions:

First, let's solve this problem by hand, using the "Z" table (see handout).

(a) What is the probability that a ball bearing will be between the target and the actual mean?

$P(0.75 < X < 0.753) =
P(\frac{0.75 - 0.753}{0.004} < Z < \frac{0.753 - 0.753}{0.004}) =$
P(-0.75 < Z < 0) = .2734

normal curve

(b) What is the probability that a ball bearing will be between the lower specification limit and the target?

$P(0.74 < X < 0.75) =
P(\frac{0.74 - 0.753}{0.004} < Z < \frac{0.75 - 0.753}{0.004}) =$
P(-3.25 < Z < -0.75) = .49942 - .2734 = .22602

normal curve

(c) What is the probability that a ball bearing will be above the upper specification limit?

$P(X > 0.76) = P(Z > \frac{0.76 - 0.753}{0.004}) =$
P(Z > 1.75) = .5 - .4599 = .0401

normal curve

(d) What is the probability that a ball bearing will be below the lower specification limit?

$P(X < 0.74) = P(Z < \frac{0.74 - 0.753}{0.004}) =$
P(Z < -3.25) = .5 - .49942 = .00058

normal curve

(Note: The shading to the left of -3.25 doesn't really show up on the picture.)

(e) Above which value in diameter will 93% of the ball bearings be?

normal curve

normal curve

The value asked for here will be the 7th percentile, since 93% of the ball bearings will have diameters above that.
So we will look up .4300 in the Z-table in a "backwards" manner.   The closest area to this is .4306, which corresponds to a Z-value of 1.48.   So the X value, once we "un-standardize", is:

$-1.48 = \frac{X - 0.753}{0.004}$
-0.00592 = X - 0.753
X = 0.74708

So 0.74708 in. is the value that 93% of the diameters are above.

Now, let's go back and get the normal probabilities with the TI-83, instead of using the Z-table.

(a) P(-.75 < Z < 0)

Since we already standardized the X values, we can use the normalcdf command (found under DISTR) with the arguments (-.75, 0).   The answer is .2733727211, which is what we got before.

TI-83 screen TI-83 screen

A nice feature of the TI-83 is the ShadeNorm function, which is found under DISTR, then DRAW.   Give it the same arguments as normalcdf.
Note: It will be necessary to set the WINDOW variables to appropriate values so the plot is visible on the screen.

TI-83 screen

(b) P(-3.25 < Z < -0.75)

TI-83 screen TI-83 screen

Note: Before you draw another normal curve, select ClrDraw from the DRAW menu.

(c) $P(Z > 1.75) = P(1.75 < Z < \infty)$

TI-83 screen TI-83 screen

Note: Since we want the probability greater than 1.75, the (lower bound, upper bound) interval is $(1.75, \infty)$. You can substitute any large number (ten would be sufficient) for infinity, but the largest number the calculator understands is 1E99, or $1 \times 10^{99}$.

(d) $P(Z < -3.25) = P(-\infty < Z < -3.25)$

TI-83 screen TI-83 screen

Note that the Z-value of -3.25 is too extreme for the shading to even show up.   This is a very small probability.

(e) The invNorm function gives the value associated with an area to the left of that value.   So for this problem, the area to be supplied is 7%.

TI-83 screen

And we get the Z-value that we obtained before, without having to use the Z-table.   It remains to be converted back to an X value, like we did before.

Of course, we can also skip doing the standardization step by hand, letting the TI-83 do all the work by specifying $\mu$ and $\sigma$ in the normalcdf command.
Let us go back one last time and see how this works.

(a) P( 0.75 < X < 0.753)

TI-83 screen TI-83 screen

Note: You will have to reset the WINDOW variables to be able to see the normal curve.

TI-83 screen

(b) P(0.74 < X < 0.75)

TI-83 screen TI-83 screen

(c) $P(X > 0.76) = P(0.76 < Z < \infty)$

(d) $P(X < 0.74) = P(-\infty < Z < 0.74)$

TI-83 screen

(e) This problem is a snap now:

TI-83 screen

Note that we didn't have to make a single hand calculation or use the Z-table. nbsp; We did, however, need to know the correct values to supply as arguments. nbsp; The best way to get those is to draw a picture (a curve) and give it some thought.   Of course, use your common sense to decide if the answer the calculator comes up with is reasonable.   Let it do the calculations for you, but not the critical thinking!



 
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Updated: 20 August 2003