Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 21, \sigma = 0.05[/tex]
a) What proportion of rods has a length less than 20.9 ​cm?
This is the pvalue of Z when X = 20.9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.9 - 21}{0.05}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.88 - 21}{0.05}[/tex]
[tex]Z = -2.4[/tex]
[tex]Z = -2.4[/tex] has a pvalue of 0.0082.
Longer than 21.12
1 subtracted by the pvalue of Z when X = 21.12
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21.12 - 21}{0.05}[/tex]
[tex]Z = 2.4[/tex]
[tex]Z = 2.4[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21.1 - 21}{0.05}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
X = 20.9
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.9 - 21}{0.05}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] 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.
