Respuesta :
The minimum score needed to be accepted in the university when it takes only top 15% performers where performace had mean 70 score and standard deviation of 8 is found being 78.32 approx.
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For this case, suppose that:
X = the score a student gets in the considered university's entrance exam.
Then, we have:
[tex]X \sim N(\mu = 70, \sigma = 8)[/tex]
Let X = x be the score such that scores bigger than x is 15% of all the scores applicants get.
That means, P(X > x) = 15% of the total area the distribution of X have under its normal distribution. Or, we can say that there is 15% chance of getting scores above X = x
Now, we can rewrite this statement as:
[tex]1 - P(X \leq x) = 0.15[/tex]
Converting X to Z (standard normal variate), we get:
[tex]1 - P(X \leq x ) = 0.15\\\\1 - P\left(Z \leq z = \dfrac{x - 70}{8} \right) =0.15\\\\P(Z \leq z = (x-70)/8) = 0.85[/tex]
The value Z = z for which the p-value is approx 0.85 is found from the z-tables as:
Thus, we get:
[tex]z = \dfrac{x-70}{8}\\x = 8z + 70\\x \approx 8 \times 1.04 + 70 = 78.32[/tex]
Thus, the minimum score needed to be accepted in the university when it takes only top 15% performers where performace had mean 70 score and standard deviation of 8 is found being 78.32 approx.
Learn more about z-score here:
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