Question

Steel rods we ma d ured with a mean length of 21 centimeter (cm) Because of variability in the manufacturing process the length of the rods are approximately normally distributed with a standard deviation of 006 om Complete parts(a) to d) (a) What proportion of rods has a length less than 20 9 cm? Round to four decimal places as needed) Any rods that we shorter than 2006 cm or longer than 21.14 cm are discarded What proportion of rods will be discarded Round to four decimal places as needed) Using the results of part (19.5000 rods are manufactured in a day, how many should the plant manager expect to discard (Use the answer from part to find this answer Round to the nearest Integer as needed) d) an order comes in for 10.000 steel rods, how many rods should the plant manager expect mandadure and 21.1 on? the orders are must be between 20.9 cm Round up to the integer) Enter your answer in each of the answers 12/18/19 11:59pm Sample Tests and quienes Samg o practice

Answer 1

As X: length of steel rods

Now,

$X\sim N(21,(0.06)^2)$

$\therefore Z = \frac{X-21}{0.06}\sim N(0,1)$

a ) $P(X<20.9)=P(\frac{X-21}{0.06}<\frac{20.9-21}{0.06})=P(Z<-1.6667)= 0.0478$

$\therefore$the proportion of steel rods less than 20.9 cm is 0.0478.

b )

$P(X<20.86)=P(\frac{X-21}{0.06}<\frac{20.86-21}{0.06})=P(Z<-2.33)= 0.0099$

  $P(X>21.14)=P(\frac{X-21}{0.06}>\frac{21.14-21}{0.06})=P(Z>2.33)= 0.0099$

Now , the proportion of rods to be discarded = 0.0099 + 0.0099 = 0.0198

c ) The probability of a rod getting discarded ( p ) = 0.0198

n = 5000

E ( X ) = np = 5000 . ( 0.0198 ) = 99

d )    $P(X<20.9)=P(\frac{X-21}{0.06}<\frac{20.9-21}{0.06})$

  

We are expected to discard 2 . ( 0.0478 ) = 0.0956 of the rods . So we are expected to keep ( 1 - 0.0956 ) = 0.9044 of the rods .

Now , E ( X ) = 10000 = np

or , n . ( 0.9044 ) = 10000

or , n = ( 10000 / 0.9044 ) = 11057.05 = 11058 ( approximately ) .