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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.
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