Question

42%3:14 PM *N 4 Due Wed 12/8/201911:59 pri Show Intro/Instructions Questions Q11Q2D/103/)Q47 vQ 5 [2/2) Q6 (2/2) Q7 (0:2) vQ 8 (2/2) >Q3U:2) 1012/2 Grade: 16/20 Prirt Verslon A campany pronures steel ads. The lengths of Lhe sleel rods are rornally distributed witi d mean of 262./-cm ond a standord devlotion of 1.7 cm. For shipment, 5 steel rods are bundled together. Find he probuuilly L'iel L'ie averaye engl'i of u randomly seloected bundle of steel rods is greater than 266.9 cm. PUA> 266.5-Lin) = Enter your answer as a number accurate to 4 derimal places. Answers obtained usirg exact 7- SLures or -scures rourided lo 3 ducimal places ore accepted. Poir ts poss blc: 2 Tis is :lerpl . Биomit LE

Answer 1

Solution :

Given that ,

mean = $\mu$ = 268

standard deviation = $\sigma$ = 1.7

P(x > 266.9 ) = 1 - p( x< 266.9)

=1- p [(x - $\mu$ ) / $\sigma$ < (266.9 - 268) /1.7 ]

=1- P(z < -0.647 )

= 1 - 0.2588= 0.7412

probability = 0.7412