Question

1) A company produces steel rods. The lengths of the steel rods
are normally distributed with a mean of 183.4-cm and a standard
deviation of 1.3-cm.

Find the probability that the length of a randomly selected steel
rod is between 179.9-cm and 180.3-cm.

P(179.9<x<180.3)=P(179.9<x<180.3)=

2) A manufacturer knows that their items have a normally
distributed length, with a mean of 6.3 inches, and standard
deviation of 0.6 inches.

If 9 items are chosen at random, what is the probability that their
mean length is less than 5.9 inches?

Answer #1

1)

for normal distribution z score =(X-μ)/σx | |

here mean= μ= | 183.4 |

std deviation =σ= | 1.3 |

probability that the length of a randomly selected steel rod is between 179.9-cm and 180.3-cm :

probability
=P(179.9<X<180.3)=P((179.9-183.4)/1.3)<Z<(180.3-183.4)/1.3)=P(-2.69<Z<-2.38)=0.0087-0.0036=0.0051 |

2)

here mean= μ= | 6.3 |

std deviation =σ= | 0.60 |

sample size =n= | 9 |

std error=σ_{x̅}=σ/√n= |
0.20000 |

probability
=P(X<5.9)=(Z<(5.9-6.3)/0.2)=P(Z<-2)=0.0228 |

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