Steel rods are manufactured with a mean length of 29 centimeter(cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed with a standard deviation of 0.07 cm. Complete parts (a) to (d)

(a) What proportion of rods has a length less than 28.9 cm?

(b) Any rods that are shorter than 28.84 cm or longer than 29.16 cm are discarded. What proportion 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?

(d) 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

28.9 cm and 29.1 cm?

c) For this case since we have a total of 5000 and we know that the proportion of discarded steel rods is 0.0221 we can find the number of discarded like this:

[tex] n = 0.0221*5000= 110.5 \approx 111[/tex]

d) Now we calculate the new probability of no discard like this:

[tex] P(\frac{28.9-29}{0.07} < Z<\frac{29.1-29}{0.07}) = P(-1.429< Z<1.429)=0.923-0.0765= 0.847[/tex]

So the new proportion of discard rods is 1-0.847 = 0.153

And the number of NO discarded from a batch of 10000 would be:

[tex] n = 10000*0.847 = 8470[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the length of steel rods of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(29,0.07)[/tex]

Where [tex]\mu=29[/tex] and [tex]\sigma=0.07[/tex]

And we want to find this probability:

[tex] P(X<28.9)[/tex]

And we can use the z score formula given by:

[tex] z= \frac{a-\mu}{\sigma}[/tex]

And if we use this formula and the normal standard distribution or excel we got:

[tex] P(X<28.9) = P(Z<\frac{28.9-29}{0.07}) = P(Z<-1.429) = 0.0765[/tex]

Part b

For this case we can find the probability that the rods would be not discarded calculating this:

[tex] P(28.84 < X< 29.16)[/tex]

Using again the z score we got:

[tex] P(\frac{28.84-29}{0.07} < Z<\frac{29.16-29}{0.07}) = P(-2.286< Z<2.286)=0.989-0.0111= 0.9779 [/tex]

So then the proportion of discarded would be 1-0.9779 =0.0221 or 2.21 %

Part c

For this case since we have a total of 5000 and we know that the proportion of discarded steel rods is 0.0221 we can find the number of discarded like this:

[tex] n = 0.0221*5000= 110.5 \approx 111[/tex]

Part d

Now we calculate the new probability of no discard like this:

[tex] P(\frac{28.9-29}{0.07} < Z<\frac{29.1-29}{0.07}) = P(-1.429< Z<1.429)=0.923-0.0765= 0.847[/tex]

So the new proportion of discard rods is 1-0.847 = 0.153

And the number of NO discarded from a batch of 10000 would be:

[tex] n = 10000*0.847 = 8470[/tex]

topeadeniran2 topeadeniran2 · Answer 2 · 2022-03-29T17:51:39+02:00

The probability shows that the proportion of rods has a length of less than 28.9 cm is 0.0475.

How to calculate the probability?

The proportion of rods has a length less than 28.9 cm will be:

= P(x<28.9) = P(z<28.9-29/0.06)

= P(z<-1.67)

= 0.0475

The proportion of rods that will be discarded will be:

= P(x<28.87 or 29.13<x)

= P(-2.167<z or z<2.167)

= 0.0302

Also, the expected value will be:

=0.0302 × 5000

= 151

Learn more about probability on:

https://brainly.com/question/25870256