Steel rods are manufactured with a mean length of 21 centimeter(cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed with a standard deviation of 0.05 cm. (

a) What proportion of rods has a length less than 20.9 cm?(Round to four decimal places as needed.)
(b) Any rods that are shorter than 20.88 cm or longer than 21.12 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 10000 steel rods, how many rods should the plant manager expect to manufacture if the order states that all rods must be between 20.9 cm and 21.1 cm?

a) 0.0228

b) 0.0164 = 1.64% of rods will be discarded.

c) The plant manager should expect to discard 82 rods.

d) The plant manager should expect to manufacture 9544 rods.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

a) What proportion of rods has a length less than 20.9 cm?

This is the pvalue of Z when X = 20.9. So

has a pvalue of 0.0228

0.0228 = 2.28% of rods have a length less than 20.9 cm.

(b) Any rods that are shorter than 20.88 cm or longer than 21.12 cm are discarded. What proportion of rods will be discarded?(

Shorter than 20.88

pvalue of Z when X = 20.88

has a pvalue of 0.0082.

Longer than 21.12

1 subtracted by the pvalue of Z when X = 21.12

has a pvalue of 0.9918

1 - 0.9918 = 0.0082

2*0.0082 = 0.0164

0.0164 = 1.64% of rods will be discarded.

(c) Using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard?

1.64% of 5000. So

0.0164*5000 = 82

The plant manager should expect to discard 82 rods.

(d) If an order comes in for 10000 steel rods, how many rods should the plant manager expect to manufacture if the order states that all rods must be between 20.9 cm and 21.1 cm?

Proportion of rods between 20.9 cm and 21.1cm is the pvalue of Z when X = 21.1 subtracted by the pvalue of Z when X = 20.9. So

X = 21.1

has a pvalue of 0.9772

X = 20.9

has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

Out of 10,000

0.9544*10000 = 9544

The plant manager should expect to manufacture 9544 rods.

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