Answer:
The area of the shaded region is 0.9082.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value, which is the area of the shaded region, is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 100, \sigma = 15, X = 120[/tex]
Area of the shaded region:
pvalue of Z
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 100}{15}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
So
The area of the shaded region is 0.9082.
The area of the shaded region which is the area to the left of the distribution curve is 0.9522
Given that :
We obtain the Zscore of the distribution :
The Area of the shaded region is the area to the left of the distribution :
Therefore, the area of the shaded region is 0.9522 which depicts the area to the left of the distribution.
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