Question

Question Help 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 0.06 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 24.88 cm or longer than 25.12 cm are discarded. What proportion of rods will be discarded? (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? (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? (Round up to the nearest integer.) Enter your answer in each of the answer boxes.

Answer 1

Normal distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 25 cm

Standard deviation = 0.06 cm

a) P(P(X < 24.9) = P(Z < (24.9 - 25)/0.06)

= P(Z < -1.67)

= 0.0475

b) P(shorter than 24.88 or longer than 25.12) = P(X < 24.88) + P(X > 25.12)

= P(X < 24.88) + 1 - P(X < 25.12)

= P(Z < (24.88 - 25)/0.06) + 1 - P(Z < (25.12 - 25)/0.06)

= P(Z < -2) + 1 - P(Z < 2)

= 0.0228 + 1 - 0.9772

= 0.0456

c) Number of rods manufactured = 5000

Number of rods expected to be discarded = 5000 x 0.0456

= 228

d) P(between 24.9 and 25.1 cm) = P(X < 25.1) - P(X < 24.9)

= P(Z < (25.1 - 25)/0.06) - P(Z < (24.9 - 25)/0.06)

= P(Z < 1.67) - P(Z < -1.67)

= 0.9525 - 0.0475

= 0.9050

Let n be the number of rods to be manufactured

n x 0.9050 = 5000

n = 5525