(a) In a manufacturing process, 4 % of steel rods made by a...

(a) In a manufacturing process, 4 % of steel rods made by a machine are out of tolerance (and therefore considered defective), the defectives occurring at random during production. If rods are packaged 100 per box, what is the probability that a given box will contain: (i) r = 0, 1, 2, 3, 4 or 5 defective rods (ii) At most, 5 defective rods? Calculate the probabilities using both the Binomial distribution and a Poisson approximation. (b) The diameter of the rods are distributed normally with mean of 2 cm and standard deviation of 0.008 cm. (i) What percentage of defectives can we expect if we set the tolerance limits at 2 ± 0.02 cm? (ii) What are the tolerance limits if 4 % of the rods are allowed to be defective?

Transcript

00:01 Now from the problem, 4 % of rods manufactured are known to be defective.

00:09 So we represent 4 % as p.

00:12 And then we're told that there are 100 rods in each box.

00:19 And for the first question, we are to determine the probability that there is 0, 1, 2, 3, 4 or 5 defective rods.

00:40 We are going to use both binomial and then person distribution.

00:54 So i'm going to divide my board into two.

01:06 So you have binomial on one side and then using person on the other side.

01:16 So using binomial, you know that the probability density function for a binomial distribution is given us p of x is equal to n combination x times times.

01:33 Times p wres to the power x times kow raised to the power n minus x so n is equal to 100 p is 0 .04 k is 1 minus p and that is 0 .96 okay so let's go ahead so when x is equal to 1 p of 1 becomes 100 combination i think the first x value is zero so we have hundred combinations 0 times 0 .04 which to the power 0 times 0 .96 to the power 100 minus 0 and this is equal to 0 .0 .169 now when x is is equal to 1 probability of finding 1 is equal to 100 combination 1 times 0 .04 to the power 1 times 0 .96 raised to the power 100 minus 1 and this is also equal to 0 .0703 and i'm simply going to compute the rest.

04:06 So when x is equal to 2 p of 2 is 2 equal to 0 .1450 when x is equal to 3 and p of 3 will be 0 .1973 when x is 4 probability that x is 4 is equal to 0 .1 994 and when x is 5 the probability that x is 5 is equal to 0 .1 994 and then when x is 5 the probability that x is 5 is equal to 0 .1 595 so we're going to do same using person distribution and for person distribution the probability this function is giving us mu which represents mean okay let's represent that in blue we have mu raised to the power x then the exponent of negative mu divided by x factorial so for this problem we need to compute the the mean value and then the mean is equal to n times p which is 100 times 0 .04 and that is simply 4.

06:20 So let's go ahead when x is equal to 0 probability probability that x is equal to 0 is giving us 4 raised to the power 0 then we have the exponent of negative 4 and we divide by 0 factorial that is equal to 0 .0183 and when x is equal to 1 the probability of finding 1 will be equal to 4 raise to the power 1 we have e to the power negative 4 divided by 1 factorial and this is also equal to 0 .073 3.

07:41 So we just compute the rest of the values.

07:43 When x is equal to 2, the probability at x is equal to 2 is equal to 0 .1465.

07:58 When x is equal to 3, p of 3 is equal to 0 .1954.

08:13 When x is 4, p of 4 is equal to 0 .1954.

08:14 When x is 4, p of 4 is equal to 0 .1954.

08:22 And then when x is 4, p of 4 is equal to 0 .0 .1954.

08:28 X x is 5 p of 5 is equal to 0 .1563 the next question we have to find the probability of finding at most 5 defective rods using both binomial and person distribution now the probability of finding at most 5 can bring us probability that x is less than or equal to 5 and this is equal to the probability that x is equal to 0 plus the probability that x is equal to 1 all the way to the probability that x is equal to 5 so we've computed for these values using both binomial and person the answer are right here so you are going to add the probabilities for both sides i'm going to extend my division down here.

10:10 So this is represents binomial.

10:24 So we have 0 .0169 plus 0 .0703 plus 0 .1450 plus 0 .1973 and then 0 .194...

Question

Please give Ace some feedback