IDNLearn.com: Where your questions meet expert advice and community insights. Our Q&A platform offers reliable and thorough answers to help you make informed decisions quickly and easily.
Sagot :
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 normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 90[/tex].
- The standard deviation is of 1 gram, hence [tex]\sigma = 1[/tex].
We want to find the probability of an item differing more than 3 grams from the mean, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3}{1}[/tex]
[tex]Z = 3[/tex]
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.
2 x 0.0013 = 0.0026
0.0026 x 100% = 0.26%
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 https://brainly.com/question/24663213
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.