Consider a value to be significantly low if its z score less than or equal to minus2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 21.9 and the standard deviation was 4.9. Identify the test scores that are significantly low or significantly high.

Step-by-step explanation:

Since we are dealing with z scores, then the distribution is a normal distribution. The formula for determining the z score is expressed as

z = (x - µ) / σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 21.9

σ = 4.9

For significantly low values, z = - 2

Therefore,

- 2 = (x - 21.9) / 4.9

- 2 * 4.9 = x - 21.9

- 9.8 = x - 21.9

x = - 9.8 + 21.9

x = 12.1

Significantly low test score = 12.1

For significantly high values, z = 2

Therefore,

2 = (x - 21.9) / 4.9

2 * 4.9 = x - 21.9

9.8 = x - 21.9

x = 9.8 + 21.9

x = 31.7

Significantly high score = 31.7