Some plastic rods are automatically cut into nominal lengths...

Some plastic rods are automatically cut into nominal lengths of 6 inches. Actual lengths are normally distributed about a mean of 6 inches and their standard deviation is 0.06 inch. What proportion of the rods exceeds tolerance limits of 5.9 inches and 6.1 inches? To what value does the standard deviation need to be reduced if 99% of the rods must be within tolerance?

Transcript

00:01 This is a normal distribution question.

00:02 So there is given the mean value, which is denoted by mu.

00:06 This is 6.

00:07 And the standard deviation, so the standard deviation, which is denoted by sigma, that was given as 0 .06.

00:14 So i can define the random variable x, which is normally distributed.

00:18 This is 6 and 0 .06.

00:21 So what proportion of roads exceeds the tolerance level? so the tolerance level, so for the first part, let me just graph and we can either.

00:30 I can easily understand what that means.

00:31 So the tolerance level, which was 5 .9, so this is the mean value.

00:37 So this is 5 .9 and the 6 .1 here.

00:41 So this is the tolerance level.

00:43 So the tolerance is between those two values.

00:45 But the question is being asked, what proportion of the roads exceed the tolerance level? so that means we need to get the area of this region and this pink region here.

00:56 How can i get easily the area of these regions? so the probability of the random variable axis is less than 5 .9 and plus the probability of x is greater than 6 .1.

01:08 So the total probability, which is 1, minus the probability of the x variable, which is between 5 .1, 5 .9 and 6 .1, which is this yellow shaded region.

01:18 So this is 5 .9 and 6 .1.

01:22 So how we can just find the value of this one here? i'm going to use the normal cdf function of the graphing the speed calculator, the lower boundary function.

01:29 5 .9, upper boundary is 6 .1, the mean value is 6 and the standard division 0 .06.

01:35 Let's get the answer.

01:36 This is 1 minus.

01:38 So press second variance.

01:39 The second option here, this is normal cdf, which is 5 .9, comma, and this is 6 .1, and the mean value is 6, and the standard division is 0 .06.

01:50 So the probability would be, which is 0 .09 and 56.

01:55 This is the probability we have for this one here.

01:58 If you want the percentage, we just multiply by 100, that would be 9 .56 % of the values are, exceeds the tolerance level.

02:08 And to what value does the standard division? so at this case, we don't know the standard division.

02:14 Let's define the standard division as sigma.

02:18 And this is reduced 99 % of the roots must be within the tolerance...