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Suppose a batch of steel rods produced at a steel plant have a mean length of 170170 millimeters, and a standard deviation of 1010 millimeters. If 299299 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by greater than 0.70.7 millimeters? Round your answer to four decimal places.

Respuesta :

Answer:

Required probability is 0.2262

Step-by-step explanation:

given data

mean length = 170 millimeters

standard deviation = 10 millimeters

sample n = 299

population mean by greater than = 0.7 millimeters

solution

we consider here batch of steel rod produced = x

probability that mean length sample that differ than the population mean greater than 0.7  required probability is

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 1 - P (  [tex]\bar {x} -\mu[/tex] ≤ 0.7 )   .................1

so we take here value

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 1 - [  P ( -0.7 ≤  [tex]\bar {x} -\mu[/tex] ≤ 0.7 ) ]

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 1 -  [  P (     [tex]\frac{-0.7}{\frac{10}{\sqrt{299}}}[/tex]    ≤   [tex]\frac{\bar {x} -\mu}{\frac{\sigma }{\sqrt{n}}}[/tex]    ≤    [tex]\frac{0.7}{\frac{10}{\sqrt{299}}}[/tex]    )  

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 1 [ P ( - 1.21  ≤  Z  ≤ 1.21 ) ]

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 1 ( P ( Z ≤ 1.21 ) - P ( Z ≤ - 1.21 ) )

we use here excel function that

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 1 - { (=NORMSDiST (1.21) - (=NORMSDiST (-1.21)  }

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 1 - ( 0.8869 - 0.1131 )

P ( [tex]\bar {x} -\mu[/tex] > 0.7  ) = 0.2262

so required probability is 0.2262

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