Answer:
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: [tex]z= \frac{x-\mu}{\sigma}[/tex]
μ= mean = 84
σ=standard deviation = 5
x= random number = 89
Putting these values, we get
[tex]z= \frac{89-84}{5}[/tex]
[tex]z= 1[/tex]
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%
Answer:
The percentage of the students scored higher than 89% is 15.87%.
Step-by-step explanation:
Given : The average test scores for a particular test in Algebra 2 was 84 with a standard deviation of 5.
To find : What percentage of the students scored higher than 89%?
Solution :
The formula to find z-score is
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where,
[tex]\mu=84[/tex] is the mean or average
[tex]\sigma=5[/tex] is the standard deviation
x=89 is the number
Substitute the value in the formula,
[tex]z=\frac{89-84}{5}[/tex]
[tex]z=\frac{5}{5}[/tex]
[tex]z=1[/tex]
Now, we have to find the percentage of the students scored higher than 89%
From the z-table the value of z at 1 is 0.8413.
Percentage of students scored higher than 89% is
[tex]P(x>89\%)=(1-0.8413)\times 100=0.1587\times 100=15.87\%[/tex]
Therefore, The percentage of the students scored higher than 89% is 15.87%.
