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zainababbas25
15.10.2019 •
Mathematics
The average test scores for a particular test in algebra 2 was 84 with a standard deviation of 5. what percentage of the students scored higher than 89%? explain your answer.
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Ответ:
Percentage of the students scored higher than 89% = 65.87%
Step-by-step explanation:
We need to find the percentage of students who scored higher than 89% given mean = 84 and standard deviation = 5.
We can find the percentage of students who scored higher than 89% by using z-score.
The formula for z-score is:![z= \frac{x-\mu}{\sigma}](/tpl/images/0322/2331/3472d.png)
μ= mean = 84
σ=standard deviation = 5
x= random number = 89
Putting these values, we get
Now, we know that we need to find students who score greater than 89%
so P(X>89) or P(X>1),
finding value of z= 1 which can be easily found using z score tables.
By looking at table we get the value of z = 0.3413
Subtracting the value of z from 1 we get
Percentage of the students scored higher than 89% = 1-0.3413 = 0.6587 or 65.87%
Ответ: