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

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

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=(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=(89-84)/(5)`

`z=(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)* 100=0.1587* 100=15.87\%`

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

Answer 2

Percentage of the students scored higher than 89% = 65.87%

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= (x-\mu)/(\sigma)`

μ= mean = 84

σ=standard deviation = 5

x= random number = 89

Putting these values, we get

`z= (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%