Y = 2/3 when x = 0.2
1 answer:
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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 and standard deviation
, the zscore of a measure X is given by:
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:
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
has a pvalue of 0.9649
X = 82
has a pvalue of 0.8186
0.9649 - 0.8186 = 0.1463
14.63% probability that a student scores between 82 and 90