Steel rods are manufactured with a mean length of 20 centimeter (cm). because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed with a standard deviation of 0.07cm(a)? what proportion of rods has a length less than 19. to four decimal places as )? any rods that are shorter than 19.83 cm or longer than 20.17? cm are discarded. what proportion of rods will be to four decimal places as )? using the results of part (b), if 5000? rods are manufactured in a day, how many should the plant manager expect to the answer from part b to find this answer. round to the nearest integer as )? 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 19.9 cm and 20.1 up to the nearest integer.)

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5

answer: graph 2

step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.