Question

1. scores on the sat form a normal distribution with $\mu = 500$ and $\sigma = 100$.

a) what is the minimum score necessary to be in the top 15% of the sat distribution?
b) find the range of values that defines the middle 80% of the distribution of sat scores.

Explanation:

Part (a)

Step 1: Find the z - score for the top 15%

We want to find the z - score such that the area to the right of it is 0.15. So the area to the left of the z - score is \(1 - 0.15=0.85\). Looking up the z - score in the standard normal distribution table (or using a calculator with a normal - distribution inverse function), the z - score corresponding to an area of 0.85 is approximately \(z = 1.04\) (using the standard normal table: \(z = 1.04\) gives an area of approximately 0.8508, which is close to 0.85).

Step 2: Use the z - score formula to find the score \(X\)

The z - score formula is \(z=\frac{X-\mu}{\sigma}\), where \(\mu = 500\), \(\sigma = 100\), and \(z = 1.04\). We can re - arrange the formula to solve for \(X\): \(X=\mu+z\sigma\).

Substitute the values: \(X = 500+1.04\times100\)

Calculate \(1.04\times100 = 104\), then \(X=500 + 104=604\)

Step 1: Find the z - scores for the middle 80%

The middle 80% of the data means that 10% of the data is in the left tail and 10% of the data is in the right tail. So we need to find the z - score for which the area to the left is 0.10 (left tail) and the z - score for which the area to the left is \(0.10 + 0.80=0.90\) (right - hand boundary of the middle 80%).

Looking up in the standard normal table:

• For the left - tail (area = 0.10), the z - score is approximately \(z=- 1.28\) (using the standard normal table, the z - score corresponding to an area of 0.1003 is - 1.28).
• For the right - tail (area = 0.90), the z - score is approximately \(z = 1.28\) (using the standard normal table, the z - score corresponding to an area of 0.8997 is 1.28).

Step 2: Use the z - score formula to find the corresponding \(X\) values

The z - score formula is \(X=\mu+z\sigma\).

For \(z=-1.28\):
\(X_1=500+(- 1.28)\times100=500 - 128 = 372\)

For \(z = 1.28\):
\(X_2=500 + 1.28\times100=500+128 = 628\)

Answer:

(a):
The minimum score necessary to be in the top 15% is approximately 604.

Part (b)