A process manufactures ball bearings whose diameters are...

A process manufactures ball bearings whose diameters are normally distributed with a mean of 2.505 cm and a standard deviation of 0.008 cm. Specifications call for the diameter to be in the interval 2.5 ± 0.01 cm. What proportion of the ball bearings will meet the specification? Use X to represent the diameter of the ball bearings. The Z-score of the lower limit X: ZL = ...................................... The Z-score of the upper limit X: ZU = ....................................... The area between ZL and 0 (which is the mean of the standard normal distribution) under the normal curve is ................................. The area between 0 and ZU under the normal curve is ....................................... The proportion of the ball bearings that will meet the specification is .......................................

Transcript

00:01 Hi, i'm david and i'm here to help me answer your question.

00:04 Now let me bring up your question here.

00:07 In this question we're going to review about the normal distribution.

00:14 Let me remind to you that if we have the x followed by the normal, with the mean and the standard deviation, and then if we think the x we minus the mean over the standard deviation, we will obtain the standard normal.

00:28 And this formula here will show us this one will be the standard normal will be the z and in this formula this quantity here will be called the z score and here we will see x it will follow by the normal when the mean equal to the 2 .505 standard division equal to the 0 .008.

00:58 Now the specification cone from the diameter to be in the interval specification will be inside the interval 2 .5 plus and minus 0 .01 so it means that it will be between the interval of the 2 .5 minus 0 .0 .0 1 equal to the 2 .49 and the 2 .51 so so therefore we want to find the z score of the lower limit for the specification.

01:39 So the lower limit will have the z score it will equal to this value 2 .49 we minus the mean 2 .505 we divide by the standard deviation and then if we do it we get equal to even by 0 .008, equal to the minus 1 .875.

02:16 And now from the upper limit, we will have the z -scar.

02:26 It will equal to this value here, 2 .51 minus 2 .505.

02:32 Evening by 0 .08, then we get equal to 0 .625.

02:48 And that will be the answer for the upper and level.

02:51 Limit now i want to find the area between the z l and zero so let me try to draw the graph here now for the z we will have the bell curve and then we have the value in the middle will be the zero here the lower limit we have equal to the minus 1 point a 75 the upper one will be the 0 .6 25 and the question ask me to find the area between the zl and the 0.

03:30 So it will be this one.

03:32 That will be the one we have to find here.

03:35 So to find that one, i will call this one, i will call this a, so the area a1, it will equal to the probability that the z, it will be smaller than greater than the 1 .a75 and we know that the half of the area equal to the 0 .5 don't forget the 0 .5 we minus left tail here.

04:08 So the left tail will be the z smaller than minus 1 .875.

04:13 To find this value we need to bring up the z table so let me come be the z table here.

04:20 I will put the table on the right here for you...

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