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Aneli [31]
1 year ago
9

Can someone help will mark brainlist

Mathematics
1 answer:
Natali [406]1 year ago
8 0

Answer:

The answer is

\frac{13}{60}

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Items produced by a manufacturing process are supposed to weigh 90 grams. The manufacturing procon
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Using the normal distribution, it is found that 0.26% of the items will either  weigh less than 87 grams or more than  93 grams.

In a <em>normal distribution</em> with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of 90 grams, hence \mu = 90.
  • The standard deviation is of 1 gram, hence \sigma = 1.

We want to find the probability of an item <u>differing more than 3 grams from the mean</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{3}{1}

Z = 3

The probability is P(|Z| > 3), which is 2 multiplied by the p-value of Z = -3.

  • Looking at the z-table, Z = -3 has a p-value of 0.0013.

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0.26% of the items will either  weigh less than 87 grams or more than  93 grams.

For more on the normal distribution, you can check brainly.com/question/24663213

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Read 2 more answers
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