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Cannot be simplified further
Answer:
0.13% of students have scored less than 45 points
Step-by-step explanation:
Problems of normally distributed samples can be 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:
About what percent of students have scored less than 45 points?
This is the pvalue of Z when X = 45. So
has a pvalue of 0.0013
0.13% of students have scored less than 45 points
Hey there Kia!
![\left[\begin{array}{ccc}40/2=20 \ ; \ 19/2=9.5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D40%2F2%3D20%20%5C%20%3B%20%5C%2019%2F2%3D9.5%5Cend%7Barray%7D%5Cright%5D%20)
Your answer should look like the following.
![\left[\begin{array}{ccc} \frac{19}{20y}= \frac{9.5}{40y^3} \end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B19%7D%7B20y%7D%3D%20%5Cfrac%7B9.5%7D%7B40y%5E3%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20)
Hope this helps you!
Your answer should look like the following.
Hope this helps you!