Reading readiness of preschoolers from an impoverished neighborhood (=20) was measured using a standardized test. Nationally, the mean on thios test for preschoolers is 30.9, with SD = 2.08.
a. children below the 30th percentile (in the bottom 30%) are in need of special assistance prior to attending school. What raw score marks the cut-off score for these children?
You can look up a normal distribution table to find that the z-score for the lower 30 percentile is z=-0.475.
Use
Score = μ+σz to get the raw score. Remember that the value of z is signed.
To find the cut-off score for children below the 30th percentile, we need to use the concept of z-scores. A z-score measures how many standard deviations away a particular score is from the mean.
First, let's calculate the z-score for the 30th percentile. The percentile can be converted to a z-score using the standard normal distribution table or by using the formula:
z = (x - mean) / standard deviation
For the 30th percentile, we know that the z-score will be negative since it's below the mean. We also know that the percentile is 30, which means 30% of the data falls below this point.
Using the formula, we plug in the values:
z = (x - 30.9) / 2.08
To solve for x (the cut-off score), rearrange the formula:
x = z * standard deviation + mean
In this case, we want to find the raw score for the cut-off point, so we'll replace z with the z-score for the 30th percentile.
Let's calculate the cut-off score:
z = -0.524 (30th percentile, negative since it's below the mean)
mean = 30.9
standard deviation = 2.08
x = -0.524 * 2.08 + 30.9
x = -1.087 + 30.9
x ≈ 29.813
Therefore, the raw score cut-off for children below the 30th percentile is approximately 29.813.