Steel rods are manufactured with a mean length of 21 centimeter​(cm). Because of variability in the manufacturing​ process, the lengths of t

Question

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?

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Maya 2 weeks 2021-09-10T07:40:56+00:00 1 Answer 0

Answers ( )

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    2021-09-10T07:42:04+00:00

    Answer:

    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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    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:

    \mu = 21, \sigma = 0.05

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

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

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{20.9 - 21}{0.05}

    Z = -2

    Z = -2 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

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{20.88 - 21}{0.05}

    Z = -2.4

    Z = -2.4 has a pvalue of 0.0082.

    Longer than 21.12

    1 subtracted by the pvalue of Z when X = 21.12

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{21.12 - 21}{0.05}

    Z = 2.4

    Z = 2.4 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

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{21.1 - 21}{0.05}

    Z = 2

    Z = 2 has a pvalue of 0.9772

    X = 20.9

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{20.9 - 21}{0.05}

    Z = -2

    Z = -2 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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