Steel rods are mandactured with a mean length of 20 centineter (oam) Because of

Question

Steel rods are mandactured with a mean length of 20 centineter (oam) Because of variability in the mianufactaring process, tho i the sods are approximal normally distributed with a standard deviation of 0.07 am Complete parts (aj to (d) (a) What proportion of rods has a length less than 19.9cm (Round to four d anal places as needed) b) Any rods that are shorlor than 19 84 c Droundtofour deanal places as needed) c) Using the resuts of part tb), 5000 rods are Use he anoaer trom part b to tnd this anwer Round to the nearest integer as neaded ) (d) If an order comes in for 10,000 steel s am or longer than 20 16 cm are discarded. What proporson of rods will be discarded? manufactured in a day, how many should the plant manager expect to dscand? rods, how many rods should the plant manager expect to manufactare if the order states that all rods must be between 19 9 and 20.1 am? (Round up to the nearest integer ) your answer in each of the answer boxes

Explanation / Answer

Solution:

This isn't binomial. In order to be binomial there must be exactly two possible outcomes (like heads/tails) and the number of experimental trials (coin tosses, for instance) must be known before you begin.

a) Get the z-score: z=(x-)/, where =mean and = standard deviation..

z = (19.9-20)/0.07 = -1.4285
Using Excel (or you can use a calculator or table), P(Z < z) = P(Z< -1.4285) = 0.0764 or 7.64%

b) Since both lengths are exactly 0.15 away from the average, we just need to do one side and then double the result.
z=(19.84-20)/0.07 = -2.2857

P(Z<z) = P(Z<-2.2857) = 10.989 = 1.11%
so double it for a 2-tail result: 2*1.11% = 2.22%

c) 2.22%*5000 = 111, or about 111 rods discarded

d) from part "a" we got that 7.64% are less than 19.9 cm. Since this is normally distributed, that also means that 7.64% are more than 20.1 cm (since both measurements are 0.1 cm either side of the mean). Thus 2*7.64% fall outside of the 19.9 cm to 20.1 cm range, or 15.28%.

Let n = number actually manufactured

n(1-discarded percentage) = 10,000
n(1-15.28%) = 10,000
n(0.8472) = 10,000
n = 11803.5883, or about 11804

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