Question

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!!! Question Help o Steel rods are manufactured with a mean length of 25 centimeter (cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed with a standard deviation of 0.07 cm. Complete parts (a) to (d) (a) What proportion of rods has a length less than 24.9 cm? (Round to four decimal places as needed.) (b) Any rods that are shorter than 24 83 cm or longer than 25.17 cm are discarded. What proportion of rods will be discarded? 0151 (Round to four decimal places as needed.) (c) Using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard? 75 (Use the answer from part b to find this answer Round to the nearest integer as needed.) (d) If an order comes in for 10,000 steel rods, how many rods should the plant manager expect to manufacture if the order states that all rods must be between 24.9 cm and 25.1 cm? 65789 (Round up to the nearest integer.)

Answer 1

Ansi ) Given M =25, o = 0.07 a ) P (x<24.9) pl XoM 24.9-25 0.07 P/2c-1.43) = 26.0764 P(x<24,83 % x 25.12) + p (x>25.17) PX224,83) x-w 21, 83-29). P X- 2597-25 0.07 0.07 P (24.2. 43) + P ( 23 24 from 0.00755+ - 0.99245= 10.0151 z-table) n = 5000 P = 0.0151 E = np = 5000+ 0.0151 = 75.5 ~ 76 176 should the plant manager expect to dascal c ) n = 10000

P (24.9 < x < 25.1) X-M < 24.9-25 0.07 25.1-25 2 0.07 C 8 -1.43 < ze +).43) P(201.13) - P (76-1.43) - 0.92364 -0.03436 0.8473 Enp - loooo xo.8 1173 18473 Irods 2 0.84 134 X-M 43-50 7 43 so m = 50 x = P[x343)=P(x- 0123-1) 1-8 (26-1 = 1-0.15866 = 8,84134 : Graph - option ③ Correct. 3 from z-table, Plze Z2-0.21) :: P(Z >02) = 0.42 = 8.42 2=0.20