Question

Items produced by a manufacturing process are supposed to weigh 90 grams. However, there is variability in the items produced, and they do not all weigh exactly 90 grams. The distribution of weights can be approximated by a normal distribution with a mean of 90 grams and a standard deviation of 1 gram. To monitor the production process, items are sampled from the manufacturing process and weighed several times throughout each day.

What proportion of all possible items will either weigh less than 89 grams or more than 91 grams? Round your answer to two decimal places.

What proportion of all possible sample means of 4 items will either weigh less than 89 grams or more than 91 grams? Round your answer to two decimal places.

Answer #1

1)

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

here mean= μ= | 90 |

std deviation =σ= | 1.0000 |

proportion of all possible items will either weigh less than 89 grams or more than 91 grams:

probability = | 1-P(89<X<91) | = | 1-P(-1<Z<1)= | 1-(0.8413-0.1587)= |
0.3174 |

2)

sample size =n= | 4 |

std error=σ_{x̅}=σ/√n= |
0.5000 |

proportion of all possible sample means of 4 items will either weigh less than 89 grams or more than 91 grams:

probability = | 1-P(89<X<91) | = | 1-P(-2<Z<2)= | 1-(0.9772-0.0228)= |
0.0456 |

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