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D Quizzes & Tests 3 13 PM × a Test-Kate Pebworth El if the automotive center do w.mathal.com/Student/PlayerTest.aspmestid» 10380221 78centerwinyes Quiz: Chapter 7 Review Quiz Kate Pebworth 8315 3-13P Submit Qua This Question: 1 pt 2017(I complete) This Quiz: 7 pts possibe Steel rods are manufactured with a mean length of 27 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 26.9 cm? (Round to four decimal places as needed.) (b) Any rods that are shorter than 26.83 cm or longer than 27.17 are dscarded what proportion of rods wil be dscarded? (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.) steel rods, how many rods should the plant manager expect to manufacture if the order states that all rods must be between (d) If an order comes in for 10,000 26.9 cm and 27.1 cm? (Round up to the nearest integer.) Enter your answer in each of the answer boxes.

Solution

Normal Distribution
Mean ( u ) =27
Standard Deviation ( sd )=0.07
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
A)
P(X < 26.9) = (26.9-27)/0.07
= -0.1/0.07= -1.4286
= P ( Z <-1.4286) From Standard Normal Table
= 0.0766                  
B)
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 9) = (9-27)/0.07
= -18/0.07= -257.1429
= P ( Z <-257.1429) From Standard Normal Table
= 0
P(X > 43) = (43-27)/0.07
= 16/0.07 = 228.5714
= P ( Z >228.571) From Standard Normal Table
= 0
P( X < 9 OR X > 43) = 0+0 = 0                  
C)
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 9) = (9-27)/0.07/ Sqrt ( 5000 )
= -18/0.001= -18182.7458
= P ( Z <-18182.7458) From Standard NOrmal Table
= 0
P(X > 43) = (43-27)/0.07/ Sqrt ( 5000 )
= 16/0.001 = 16162.4407
= P ( Z >16162.4407) From Standard Normal Table
= 0
P( X < 9 OR X > 43) = 0+0 = 0                  
D)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 26.9) = (26.9-27)/0.07/ Sqrt ( 10000 )
= -0.1/0.0007
= -142.8571
= P ( Z <-142.8571) From Standard Normal Table
= 0
P(X < 27.1) = (27.1-27)/0.07/ Sqrt ( 10000 )
= 0.1/0.0007 = 142.8571
= P ( Z <142.8571) From Standard Normal Table
= 1
P(26.9 < X < 27.1) = 1-0 = 1