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 0.08 cm. Click the icon to view a table of areas under the normal curve. (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 20.81 cm or longer than 21.19 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 10,000 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? FO ? 51 Enter your answer in each of the answer boxes. ada varia

Answer 1

Given data :

mean=21 cm

Std dev=0.08 cm

Z score:

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

a)What proportion of rod less than 20.9

$P(X<20.9)=P(Z<\frac{20.9-21}{0.08})$

$=P(Z<-1.25)$

  $=0.1056$ (using std normal ( Z) table )

b)prob of rod shorter than 20.81 and longer than 21.19

$=P(Z<\frac{20.81-21}{0.08})+P(Z>\frac{21.19-21}{0.08})$

$=P(Z<-2.375)+P(Z>2.375)$

$=0.0088+(1-0.9912)$ (using std normal ( Z) table )

$=0.0088+0.0088$

$=0.0176$

(c) if 5000 rod are manufactured in a day using part (b) prob

n=5000 , p=0.0176

Number of plant expect to discards

=n*p=5000*0.0176 = 88

(d)

P(20.9<X<21.1)

$=P(\frac{20.9-21}{0.08}<Z<\frac{21.1-21}{0.08})$

$=P(-1.25<Z<1.25)$

$=P(Z<1.25)-P(Z<-1.25)$

$=0.8944-0.1056$

$=0.7888$   (using std normal ( Z) table )

If n=10000 then number of rod

=n*p=10000*0.7888

=7888