# Normal Distribution

Consider exercise 6.68 on page 408 of your textbook.

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 in. and in.
For this problem, note that "Target" = .75, and "Actual mean" = .753.

• By hand (using the "Z" table)
• With the TI-83 (standardizing by hand)
• With the TI-83 (specifying and )

First, let's solve this problem by hand, using the "Z" table (page E-4 at the back of the textbook).

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

P(-0.75 < Z < 0) = .2734

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

P(-3.25 < Z < -0.75) = .49942 - .2734 = .22602

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

P(Z > 1.75) = .5 - .4599 = .0401

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

P(Z < -3.25) = .5 - .49942 = .00058

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

-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.

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.

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

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

(c)

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

(d)

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%.

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 and in the normalcdf command.
Let's go back one last time and see how this works.

(a) P( 0.75 < X < 0.753)

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

(b) P(0.74 < X < 0.75)

(c)

(d)

(e) This problem is a snap now:

Note that we didn't have to make a single hand calculation or use the Z-table. We did, however, need to know the correct values to supply as arguments. 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!