Let the event ![A=\{ Horse \}, B=\{ Female \}](https://tex.z-dn.net/?f=%20A%3D%5C%7B%20Horse%20%5C%7D%2C%20B%3D%5C%7B%20Female%20%5C%7D%20)
The the probability of the events
![P(A \cap B)=\frac{2}{14+26} =\frac{1}{20}](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20B%29%3D%5Cfrac%7B2%7D%7B14%2B26%7D%20%3D%5Cfrac%7B1%7D%7B20%7D%20%20)
![P( B)=\frac{26}{14+26} =\frac{13}{20}](https://tex.z-dn.net/?f=%20P%28%20B%29%3D%5Cfrac%7B26%7D%7B14%2B26%7D%20%3D%5Cfrac%7B13%7D%7B20%7D%20%20)
The conditional probability
![P(Horse|Female)=P(A|B)=\frac{P(A \cap B)}{P(B)} =\frac{1/20}{13/20} =\frac{1}{13}](https://tex.z-dn.net/?f=%20P%28Horse%7CFemale%29%3DP%28A%7CB%29%3D%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28B%29%7D%20%3D%5Cfrac%7B1%2F20%7D%7B13%2F20%7D%20%3C%2Fp%3E%3Cp%3E%3D%5Cfrac%7B1%7D%7B13%7D%20%20)
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.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- 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
.
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%20150%2C%20%5Csigma%20%3D%209%2C%20n%20%3D%2010%2C%20s%20%3D%20%5Cfrac%7B9%7D%7B%5Csqrt%7B10%7D%7D%20%3D%202.85)
The probability is <u>one subtracted by the p-value of Z when X = 157</u>, hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
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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so sorry not sure how to do this ! :c
Answer:
where is model A
Step-by-step explanation: