Question

What is the probability that the mean score for 10 randomly selected people who took the LSAT would be above 157? Round your answer to three decimal places. (Example: 0.398)

Answer 1

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 $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• 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 $s = \frac{\sigma}{\sqrt{n}}$.

Researching this problem on the internet, the parameters are given as follows:

$\mu = 150, \sigma = 9, n = 10, s = \frac{9}{\sqrt{10}} = 2.85$

The probability is one subtracted by the p-value of Z when X = 157, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

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 brainly.com/question/15181104

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