a) 88.10% of students would be expected to score over 80
b) 25.78% of students would be expected to score under 89.
c) 22.34% of students would be expected to score between 111 and 131.
d) 4.95% of students would be expected to score over 128.
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:
a) over 80?
1 subtracted by the pvalue of Z when X = 80.
has a pvalue of 0.1190
1 - 0.1190 = 0.8810
88.10% of students would be expected to score over 80
b) under 89?
Pvalue of Z when X = 89. So
has a pvalue of 0.2578.
25.78% of students would be expected to score under 89.
c) between 111 and 131?
Pvalue of Z when X = 131 subtracted by the pvalue of Z when X = 111.
X = 131
has a pvalue of 0.9656.
X = 111
has a pvalue of 0.7422.
0.9656 - 0.7422 = 0.2234
22.34% of students would be expected to score between 111 and 131.
d) over 128?
1 subtracted by the pvalue of Z when X = 128. So
has a pvalue of 0.9505
1 - 0.9505 = 0.0495
4.95% of students would be expected to score over 128.
