Answer:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And we can solve for the value of X like this:
[tex] X = \mu + z*\sigma[/tex]
And since we know that z=2 we can replace and we got:
[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we can assume the distribution for X is given by:
[tex]X \sim N(49,20.7)[/tex]
Where [tex]\mu=49[/tex] and [tex]\sigma=20.7[/tex]
And for this case the z score is given by:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And we can solve for the value of X like this:
[tex] X = \mu + z*\sigma[/tex]
And since we know that z=2 we can replace and we got:
[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]