Answer:
6.18% of the class has an exam score of A- or higher.
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:
What percentage of the class has an exam score of A- or higher (defined as at least 90)?
This is 1 subtracted by the pvalue of Z when X = 90. So
has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% of the class has an exam score of A- or higher.
Answer:
Step-by-step explanation:
I got you kiddo
You do 2*d - 2*5 = 8 which is 2d-10=8 ------> 2d=8+10 and you do d=18/2 which d = 9
Answer:
35% of 80 is 28.
Step-by-step explanation:
Answer:
D is -31 also i checked to make sure.