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Sagot :
Using the normal distribution, there is a 0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.
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.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
Researching this problem on the internet, the parameters are given as follows:
[tex]\mu = 150, \sigma = 9, n = 10, s = \frac{9}{\sqrt{10}} = 2.85[/tex]
The probability is one subtracted by the p-value of Z when X = 157, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (157 - 150)/2.85
Z = 2.46
Z = 2.46 has a p-value of 0.993.
1 - 0.993 = 0.007.
0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.
More can be learned about the normal distribution at https://brainly.com/question/15181104
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