Answer :
Answer:
14.63% probability that a student scores between 82 and 90
Step-by-step explanation:
Problems of normally distributed samples are 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 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 = 72.3, \sigma = 8.9[/tex]
What is the probability that a student scores between 82 and 90?
This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So
X = 90
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{90 - 73.9}{8.9}[/tex]
[tex]Z = 1.81[/tex]
[tex]Z = 1.81[/tex] has a pvalue of 0.9649
X = 82
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{82 - 73.9}{8.9}[/tex]
[tex]Z = 0.91[/tex]
[tex]Z = 0.91[/tex] has a pvalue of 0.8186
0.9649 - 0.8186 = 0.1463
14.63% probability that a student scores between 82 and 90