Question Help rods are manufactured with a mean length of 24 centimeter (cm), Be

Question

Question Help rods are manufactured with a mean length of 24 centimeter (cm), Because of variability in the manufacturing process, the lengths of the rods are approximately nomally distributed with a standard deviation of .06 cm. Click the ion to view a table of areas under the normal orve. )Whatproportion of rods has alongh less than 23.9cm? (Round to four decimal paces as needed ) (b) Any rods that are shorter than 23.88 cm or longer than 24.14 cm are discarded. What proportion of rods will b Round to four decimal places as needed.) e) usigthne results of pan (b),500 reds are manufactured in a day, how many should to plant manager eaged to discard? 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 Rand up to the nearest integer ) many rods shouis the piant manager expect to manufacture if the order states that all nods must be between 23.9 om and 24.1 m

Explanation / Answer

Solution:

Let X be length of rod

Then, X ~ N(21, 0.07²)

Normalising we get, Z ~ (0, 1), where P(X < x) = P(Z < (x - 24)/0.06)

(a) P(X < 23.9) => P(Z < (23.9 - 24)/0.06) => P(Z < -1.67)
1P ( Z<1.67 )=10.9525=0.0475

(b) P(Z < (23.86 - 24)/0.06) and P(24.14- 24)/0.06) to be dicarded.

=> P(Z < -2.33) and P(Z > 2.33)

From tables: (10.9901) + (0.9901) = 0.9802

(c) 5000 x 0.9802 = 4901...so, could expect 9 or 10 to be discarded

(d) P(Z < 23.9) = 0.0475

P(Z > (24.1 - 24)/0.06) = P(Z > 1.67) = 0.0475

i.e. rejection proportion is 0.0475 + 0.0475 = 0.095

=> acceptance proportion is 0.905

so, we require that 0.905 x Q = 10,000

so, Q = 10,000/0.905 = 11049.7238

=> the manager needs to manufacture 11049.72

N.B. Is the planet manager really the plant manager.

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