# Probability And Statistics - Question

• Aug 15th 2006, 11:44 AM
c00ky
Probability And Statistics - Question
Hi guys,

Problem:

The mean diameter of a batch of bolts is 9.711mm and the standard deviation of the batch is 0.126mm

The tolerance for this batch of components is 9.73mm to 9.97mm.
In a batch of 2000 bolts, determine the following:

a) The percentage of bolts rejected assuming the bolts are normally distributed.

b) The number of bolts greater than 9.91mm

c) The number of bolts less than 9.4mm

It's a tricky one!
• Aug 15th 2006, 12:39 PM
galactus
part b.

P(x>9.91)

$\frac{9.91-9.711}{0.126}=1.58$

1.58 corresponds to 0.9429

1-0.9429=0.0571

A little less than 6% are greater than 9.91mm

.0571(2000)=114.2, rounded to 115, are greater than 9.91mm

Do the same thing on the others.
• Aug 15th 2006, 12:58 PM
c00ky
i'm confused.. could you talk me through your solution?
• Aug 15th 2006, 01:52 PM
galactus
Since you want to know the number of bolts greater than 9.91 mm, you must figure the z-score and subtract it from 1. Because it is the area to the right of 9.91. See?.

Draw a normal curve, label the mean in the center and 9.91 to the right of that. Color the area to the right of 9.91. That is the area you want.

For part a, you do the same type of thing. Since the tolerance is between 9.73 and 9.97. Anything over 9.97 or under 9.73 is rejected. Find the area to the right of 9.97 and to the left of 9.73, then add them. That's the percentage you want.

part c, find the area to the left of 9.4. Because that'll be the number of bolts less than 9.4 mm.

$\frac{9.4-9.711}{0.126}\approx{-2.47}$

Look up -2.47 in the table. You find 0.0068 .

.0068(2000)=13.6. About 14 are less than 9.4 mm.

What that means is about 14 bolts will be 2.47 standard deviations below the mean.

Note: The values in the standard normal distribution table are from
${-\infty}$ to z. Therefore, if you want an area to the right you must subtract from 1.

Did you ever wonder where those values in the table come from?. They're the area under the normal curve. Therefore, an integral is used to find them.

$\frac{1}{\sqrt{2{\pi}}}\int_{-\infty}^{x}e^{\frac{-t^{2}}{2}}dt$

I know you don't need to know that, but if you're curious.

Do you have a textbook?. Any good stats book will describe this in detail.