Reading readiness of preschoolers from an impoverished neighborhood (n = 20) was measured using a standardized test. Nationally, the mean on this test for preschoolers is 30.9, with s = 2.08.
a.Children below the 30th percentile are in need of special assistance prior to attending school. What score would indicate these children?
b.What percentage of children score between 25 and 28.5?
c.How many children would we expect to find with scores between 28 and 31.5?
d.Children in the top 25% are considered accelerated readers and qualifydifferent placement in school. What score would mark the cutoff for such placement?
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To answer these questions, we need to use the concept of z-scores. A z-score measures how many standard deviations a particular score is away from the mean. It helps us compare scores from different distributions.
First, let's calculate the z-score for each scenario using the formula:
z = (x - μ) / σ
Where:
x = the score we want to determine
μ = the mean of the distribution
σ = the standard deviation of the distribution
a. Children below the 30th percentile:
To find the score that indicates children below the 30th percentile, we need to find the z-score corresponding to the 30th percentile, which is -0.52 (approximately).
Using the formula, we can rearrange it to solve for x:
x = z * σ + μ
x = -0.52 * 2.08 + 30.9
x ≈ 29.2
Therefore, a score of approximately 29.2 or below would indicate these children.
b. Percentage of children scoring between 25 and 28.5:
To find the percentage of children who scored between 25 and 28.5, we need to calculate the z-scores for both values and then find the area between those z-scores on the normal distribution curve.
For 25:
z1 = (x - μ) / σ
z1 = (25 - 30.9) / 2.08
z1 ≈ -2.84
For 28.5:
z2 = (x - μ) / σ
z2 = (28.5 - 30.9) / 2.08
z2 ≈ -1.15
Using a standard normal table or a statistical software, you can find the area between these z-scores. Let's assume the area is A. The percentage of children scoring between 25 and 28.5 is approximately A * 100%.
c. Number of children expected to score between 28 and 31.5:
Similar to part b, we find the z-scores for 28 and 31.5, and then find the area between them on the normal distribution curve.
For 28:
z1 = (28 - 30.9) / 2.08
z1 ≈ -1.39
For 31.5:
z2 = (31.5 - 30.9) / 2.08
z2 ≈ 0.29
Using the area A between these z-scores, we can multiply it by the sample size (n = 20) to estimate the number of children expected to score between 28 and 31.5:
Number of children = A * n
d. Score marking the cutoff for top 25%:
To find the score that marks the cutoff for the top 25%, we need to find the z-score that corresponds to the 75th percentile. This is because the top 25% corresponds to the upper portion of the distribution.
Using a standard normal table or a statistical software, look up the z-score corresponding to the 75th percentile. Let's assume it is z = 0.67 (approximately).
Now, rearrange the formula to solve for x:
x = z * σ + μ
x = 0.67 * 2.08 + 30.9
x ≈ 32.6
Therefore, a score of approximately 32.6 or above would mark the cutoff for such placement.