Respuesta :
Using the normal distribution, it is found that there is a 0.0228 probability that a randomly selected score is greater than 334.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are, respectively, given by [tex]\mu = 310, \sigma = 12[/tex]
The probability that a randomly selected score is greater than 334 is one subtracted by the p-value of Z when X = 334, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{334 - 310}{12}[/tex]
Z = 2
Z = 2 has a p-value of 0.9772
1 - 0.9772 = 0.0228.
0.0228 probability that a randomly selected score is greater than 334.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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