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Sagot :
Using the normal distribution, it is found that she scores less than 128 in 28.1% of her games.
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 given, respectively, by [tex]\mu = 135, \sigma = 12[/tex].
The proportion of games in which she scores less than 128 is the p-value of Z when X = 128, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{128 - 135}{12}[/tex]
Z = -0.58
Z = -0.58 has a p-value of 0.281.
She scores less than 128 in 28.1% of her games.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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