Question

6 A factory manufactures pencils and the plastic boxes that they are stored in. The lengths of the pencils made in the factory are normally distributed with mean 18 cm and standard deviation 0.1 cm. blank Leave The plastic boxes used to store the pencils are 18.2 cm long. 5 random pencils are put into each box. If any of the pencils do not fit in the box, the box will be damaged. Find the probability that a box will be damaged. a) You may assume that the sides of the box have negligible thickness. b) The factory produces 1000 boxes of pencils each day. Using a normal approximation, estimate the probability that more than 125 boxes will be damaged in a single day.

Answer 1

a) mean: 1-18 cm, and standard deviation ơ-0.1 cm. The plastic boxes used to store the pencils are 18.2 cm long. 5 random pencils are put into each box. If any of the pencils do not fit in the box, the box will be damaged. a) n, a it is a small number, e use t-distribution, and so the t-score for the 18.2 cm, is given by 18.2-џ 18.2-18 2V5 4.472 degree of freedomn-1-5-1-4 By looking up the t-distribution table for degree of freedom 4, we find As the probability for the t -score less than or equal to 4.472 is 0.9945, thus the probability that a box will be damaged is 1-0.99450.0055

b) The factory produces 1000 boxes of pencils each day. Using a normal approximation, estimate the probability that more than 125 boxes will be damaged in a single day when 5 random pencils are put into each box the probability that a box will be damaged is 0.0055, so when 1000 boxes of pencils are produced Here we use normal approximation to binomial distribution. We consider success as a box is damaged. Then its probability is p 0.0055, 1 -p0.9945 Thus q Also number of boxes-n 1000 Now np 1000 x 0.0055 5.5 nq 1000 x 0.9945 994.5 As both are greater than 5, so we can use normal approximation. Now mean-np-5.5 Standard deviation-Vnpq v 1000 x 0.0055 0.9945 2.3387

Now to find that more than 125 boxes will be damaged in a single day. That is P(X 2 125) Using the continuity correction factor P(X 2 125) -P(X> 125 - 0.5)-P(X > 124.5) Now the z-score 124.5-55 2.3387 50.88 As the z-score is too high, so the probability is O