Question

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%? Please explain your answer.

Answer 1

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: $z= \frac{x-\mu}{\sigma}$

μ= mean = 84

σ=standard deviation = 5

x= random number = 89

Putting these values, we get

$z= \frac{89-84}{5}$

$z= 1$

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 2

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  

$z=\frac{x-\mu}{\sigma}$

Where,

$\mu=84$ is the mean or average

$\sigma=5$ is the standard deviation

x=89 is the number

Substitute the value in the formula,

$z=\frac{89-84}{5}$

$z=\frac{5}{5}$

$z=1$  

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

$P(x>89\%)=(1-0.8413)\times 100=0.1587\times 100=15.87\%$

Therefore, The percentage of the students scored higher than 89% is 15.87%.