Explore a vast range of topics and get informed answers at IDNLearn.com. Ask anything and get well-informed, reliable answers from our knowledgeable community members.
Sagot :
Answer:
0.1131 = 11.31% probability that a randomly selected stock will close up $0.75 or more.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with a mean of $0.35 and a standard deviation of $0.33.
This means that [tex]\mu = 0.35, \sigma = 0.33[/tex].
What is the probability that a randomly selected stock will close up $0.75 or more?
This is 1 subtracted by the p-value of Z when X = 0.75. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.75 - 0.35}{0.33}[/tex]
[tex]Z = 1.21[/tex]
[tex]Z = 1.21[/tex] has a p-value of 0.8869.
1 - 0.8869 = 0.1131
0.1131 = 11.31% probability that a randomly selected stock will close up $0.75 or more.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the answers you need. Thank you for visiting, and we look forward to helping you again soon.
