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25.8 is the probability that a randomly selected score is greater than 334. What i did was i divided 334 by 12 and i got 25.833333333 but the real answer is 25.8. I hope i could help!!
The probability that a randomly selected score is greater than 334 is 0.0228
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, we're given that:
- Combined scores were approximately normally distributed with mean of 310 and a standard deviation of 12.
If we take:
X = a randomly selected score obtained in the GRE exam.
Then, by the given data, we have:
[tex]X \sim N(\mu = 310, \sigma = 12)[/tex]
And therefore, we can write:
Probability that a randomly selected score is greater than 334 = P(X > 334)
Converting X to standard normal distribution:
[tex]Z = \dfrac{X - \mu}{\sigma} = \dfrac{X - 310}{12}[/tex]
The probability we need can be calculated as:
[tex]P(X > 334) = 1 - P(X \leq 334)\\P(X > 334) = 1 - P(Z \leq \dfrac{334 - 310}{12})\\\\P(X > 334) = 1 - P(Z \leq 2)[/tex]
From the z-tables, the p-value for Z = 2 is 0.9772
Thus, we get [tex]P(Z \leq 2) = 0.9772[/tex], and therefore,
[tex]P(X > 334) = 1 - P(Z \leq 2) = 1- 0.9772 = 0.0228[/tex]
Thus, the probability that a randomly selected score is greater than 334 is 0.0228
Learn more about z-score here:
https://brainly.com/question/21262765
