Question

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?

Answer 1

Answer:

a) $P(X$

b) $P(\frac{28.84-29}{0.07} < Z$

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

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:

$n = 0.0221*5000= 110.5 \approx 111$

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

$P(\frac{28.9-29}{0.07} < Z$

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:

$n = 10000*0.847 = 8470$

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:

$X \sim N(29,0.07)$  

Where $\mu=29$ and $\sigma=0.07$

And we want to find this probability:

$P(X$

And we can use the z score formula given by:

$z= \frac{a-\mu}{\sigma}$

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

$P(X$

Part b

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

$P(28.84 < X< 29.16)$

Using again the z score we got:

$P(\frac{28.84-29}{0.07} < Z$

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:

$n = 0.0221*5000= 110.5 \approx 111$

Part d

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

$P(\frac{28.9-29}{0.07} < Z$

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:

$n = 10000*0.847 = 8470$