Homework 4: Normal Curve And Z Scores — 70 Points Possible

Homework 4normal Curve And Z Scores 70 Pts Possiblethis Homework Req

Homework 4 normal curve and z scores (70 pts possible) This homework requires both your text and your calculator. The objective of your fourth homework assignment involves answering questions related to the normal and standard normal curves. You will need to use the “Normal Curve Table” in Table A-1 of the Appendix of your text, as discussed in Chapter 3. You will submit this assignment as a Word document, showing your work for each problem. The assignment includes questions about the distribution of preschoolers' test scores and age at onset of dementia, requiring calculations of raw scores, percentages, expected counts, standard deviations, and cutoff points based on z-scores.

Paper For Above instruction

The following paper provides a comprehensive response to the homework assignment centered on the normal distribution, z-scores, and their application in psychological and educational research contexts. It demonstrates knowledge of statistical concepts, calculations, and interpretation of results related to the normal curve, based on the specific questions laid out in the assignment.

Introduction

The normal distribution, also known as the Gaussian distribution, is a fundamental concept in statistics, especially in psychology and education research. It describes how the values of a variable are distributed, with most observations clustering around the mean and fewer in the extremes. The standard normal distribution is a special case with a mean of 0 and a standard deviation of 1, which allows for the use of z-scores to interpret raw data relative to the mean. By calculating z-scores and consulting standard normal tables, we can determine the percentage of the population that falls within specific ranges, identify cut-off scores, and predict expected frequencies.

Question 1: Preschool Test Scores and Percentile Cut-offs

a. Children below the 30th percentile (bottom 30%)—raw score cut-off

To find the raw score corresponding to the 30th percentile, we first identify the z-score associated with the cumulative area of 0.30 in the standard normal table. Consulting Table A-1, the z-score for 0.30 is approximately -0.52. Using the z-score formula:

z = (X - μ) / σ

Rearranged to solve for X:

X = z * σ + μ

Substituting known values:

X = -0.52 * 2.08 + 30.9 ≈ -1.082 + 30.9 ≈ 29.82

Therefore, the raw score cutoff for children below the 30th percentile is approximately 29.82.

b. Percentage of children scoring between 25 and 28.5

First, convert the raw scores to z-scores:

• X = 25: z = (25 - 30.9) / 2.08 ≈ -5.9 / 2.08 ≈ -2.84
• X = 28.5: z = (28.5 - 30.9) / 2.08 ≈ -2.4 / 2.08 ≈ -1.15

Consulting the standard normal table, the area to the left of z = -2.84 is approximately 0.0023, and z = -1.15 corresponds to approximately 0.1251. Therefore, the percentage of children scoring between these scores is:

0.1251 - 0.0023 = 0.1228 or 12.28%

c. Number of children scoring between 28 and 31.5

• X = 28: z ≈ -1.15 (as above)
• X = 31.5: z = (31.5 - 30.9) / 2.08 ≈ 0.6 / 2.08 ≈ 0.29

The area to the left of z = 0.29 is approximately 0.6141. The area to the left of z = -1.15 is 0.1251. Thus, the proportion scoring between 28 and 31.5 is:

0.6141 - 0.1251 = 0.489 or 48.9%

Given n = 20 children, the expected number of children in this range is:

20 * 0.489 ≈ 9.78 ≈ 10 children

d. Raw score for top 25% (children considered accelerated readers)

Finding the z-score for the 75th percentile (since the top 25% starts above this value):

The z-score corresponding to an area of 0.75 is approximately 0.674.

Calculating raw score:

X = 0.674 * 2.08 + 30.9 ≈ 1.403 + 30.9 ≈ 32.30

Thus, children with scores above approximately 32.30 are in the top 25%.

Question 2: Age at Onset of Dementia

a. Mean and standard deviation

Given data: ΣX = 1008, Σ(X - M)^2 = 140.4, and sample size n=15.

The mean (M) is: M = ΣX / n = 1008 / 15 ≈ 67.2 years

The variance (s^2): s^2 = Σ(X - M)^2 / (n - 1) = 140.4 / 14 ≈ 10.03

The standard deviation (s): s = √10.03 ≈ 3.17 years

b. Percentage of individuals showing signs at or before age 62

Calculate the z-score for age 62:

z = (62 - 67.2) / 3.17 ≈ -5.2 / 3.17 ≈ -1.64

From the standard normal table, the area to the left of z = -1.64 is approximately 0.0505, meaning about 5.05% of the individuals might show signs of dementia at or before age 62.

c. Ages corresponding to ±1 z-score from the mean

Lower age: M - 1 * s = 67.2 - 3.17 ≈ 64.03 years

Upper age: M + 1 * s = 67.2 + 3.17 ≈ 70.37 years

Therefore, the normal range spans approximately from 64.03 to 70.37 years.

d. Ages marking the top and bottom 10% boundaries

The z-score for the top 10% (90th percentile) is approximately 1.28, and for the bottom 10% (10th percentile) is -1.28.

Calculations:

• Top 10%: X = 1.28 * 3.17 + 67.2 ≈ 4.05 + 67.2 ≈ 71.25 years
• Bottom 10%: X = -1.28 * 3.17 + 67.2 ≈ -4.05 + 67.2 ≈ 63.15 years

Thus, the most deviant individuals are those below approximately 63.15 years or above approximately 71.25 years.

Conclusion

This analysis illustrates how z-scores and the normal distribution are vital for understanding variations within populations. In educational settings, they help identify students requiring additional support, while in clinical research, they assist in characterizing age-related phenomena such as dementia onset. Accurate calculations of raw scores, expected frequencies, and cutoff points empower practitioners and researchers to make informed decisions based on statistical evidence, emphasizing the importance of mastering normal curve concepts in psychological and health sciences.

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