Question

In 1994 , the average score for the Verbal SAT was 500 with SD of 110. Determine the following : (a) the percentage of students that scored 600 or more on the Verbal SAT (b) the percentage of students that scored 378 or less on the Verbal SAT, and () the percentage of students that scored between 450 and 600 on the Verbal SAT.

Answer 1

The scores of the SAT will be assumed to be normally distributed.

According to the data provided:

`\begin{gathered} \mu=500\text{ (The average)} \\ \sigma=110\text{ (The standard deviation)} \end{gathered}`

(a). The percentage of students that scored 600 or more.

To estimate that we need to calculate the z-value corresponding to 600. The following equation is used:

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

Where x is the value (600 in this case). Replacing values:

`z=(600-500)/(110)=0.909`

We can look for this value in a normal distribution table. The normal distribution table will give us the percentage of the students that scored below 600. In the end, we subtract that value from 1 to obtain the percentage who scored higher than that. We will need to interpolate.

The area corresponding to z = 0.9 is 0.81594, while the area corresponding to z = 0.91 is 0.81859. Interpolating the values, wi will get a value of 0,81832 for z = 0.909. Then, the percentage scoring below 600 is 81.83%.

The percentage scoring above 600 will be:

`100\text{\%}-81.83\text{\%}=18.17\text{\%}`

Answer (a): 18.17% of the students scored 600 or more.

(b). To calculate the students scoring below 378 we can follow the same process. Finding the z-value we can look in the table for the corresponding area at the left of that z-value, which will represent the fraction of students below that score.

`z=(x-\mu)/(\sigma)=(378-500)/(110)=-1.109`

The corresponding z-value is -1.109.

From the table, the area corresponding to z = -1.11 is 0..13350, while the one corresponding to z = -1.1 is 0.13567. Interpolating values, the area corresponding to z = -1.109 is 0.1337, that is, 13.37%.

Answer (b): 13.37% of the students scored 378 or less.

(c). Tho find the percentage of students scoring between 450 and 600 we will need to find the areas at the left of both 600 and 450, and then subtract them. From point (a), we know that the area at the left of 600 is 81.83%. We just need to find the area at the left of 450. Let's calculate the z-value:

`z=(x-\mu)/(\sigma)=(450-500)/(110)=-0.454`

From the table, the area corresponding to z = -0.46 is 0.32276, while the one corresponding to z = -0.45 is 0.32636. Interpolating for z = -0.454 we obtain a value of 0.3249, or well, 32.49%. To find the area between 450 and 600 we subtract the area at the left of 450 to the area at the left of 600:

`81.83\text{\%}-32.49\text{\%}=49.34\text{\%}`

Bear in mind that the area at the left of a certain value represents the fraction of students scoring below that value.

The area between 450 and 600 is 49.34%.

Answer (c): 49.34% of the students scored between 450 and 600.