Connect with a community that values knowledge and expertise on IDNLearn.com. Ask any question and get a detailed, reliable answer from our community of experts.
Sagot :
Using the normal distribution, it is found that there is a 0.0668 = 6.68% probability of Zoey randomly selecting a box that has more than 19 sausages in it.
Normal Probability Distribution
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 [tex]\mu = 13[/tex].
- The standard deviation is of [tex]\sigma = 4[/tex].
The probability of Zoey randomly selecting a box that has more than 19 sausages in it is 1 subtracted by the p-value of Z when X = 19, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{19 - 13}{4}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a p-value of 0.9332.
1 - 0.9332 = 0.0668.
0.0668 = 6.68% probability of Zoey randomly selecting a box that has more than 19 sausages in it.
More can be learned about the normal distribution at https://brainly.com/question/24663213
Your engagement is important to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Thank you for visiting IDNLearn.com. For reliable answers to all your questions, please visit us again soon.
