Age at onset of dementia was determined for a sample of adults between the ages of 60 and 75. For 15 subjects, the results were SumX=1008, and Sum(X-M)^2=140.4. Use this information to answer the following:
a. What is the mean and SD for this data?
b. Based on the data you have and the Normal Curve Tables, what percentage of people might start to show signs of dementia at or before age 62?
c. If we consider the normal range of onset in this population to be +/-1 z-score from the mean, what two ages correspond to this?
d. A neuropsychologist is interested only in studying the most deviant portion of this population, that is, those individuals who fall within the top 10% and the bottom 10% of the distribution. She must determine the ages that mark these boundaries. What are these ages?
a. Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
b. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
c. Z = (score-mean)/SD
d. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.10) related to the Z score.
I'll let you do the calculations.
I found the mean but i do not know what the scores are. help. thanks
a. To calculate the mean and standard deviation for this dataset, we need to use the given information:
SumX = 1008 (sum of all the ages)
Sum(X-M)^2 = 140.4 (sum of the squared deviations from the mean)
To find the mean (M), we divide the sum of ages (SumX) by the number of subjects (15):
M = SumX / n
M = 1008 / 15
M ≈ 67.2
To find the standard deviation (SD), we need to take the square root of the variance:
Variance = Sum(X-M)^2 / (n-1)
Variance = 140.4 / (15-1)
Variance ≈ 9.36
SD = sqrt(Variance)
SD = sqrt(9.36)
SD ≈ 3.06
Therefore, the mean is approximately 67.2 and the standard deviation is approximately 3.06.
b. To determine the percentage of people who might start to show signs of dementia at or before age 62, we can use the Normal Curve Tables. However, to use the tables, we first need to standardize the value using z-scores.
The formula for calculating the z-score is: z = (X - M) / SD
For X = 62:
z = (62 - 67.2) / 3.06
z ≈ -1.69
Now, we can use the Normal Curve Tables to find the percentage corresponding to this z-score. Looking up the value -1.69 in the table will give us the percentage of people who might start to show signs of dementia at or before age 62.
c. To find the two ages corresponding to the normal range of onset in this population (+/-1 z-score from the mean), we can use the z-score formula:
For z = +1:
X = M + (z * SD)
X = 67.2 + (1 * 3.06)
X ≈ 70.26
For z = -1:
X = M + (z * SD)
X = 67.2 + (-1 * 3.06)
X ≈ 64.14
Therefore, the two ages corresponding to the normal range of onset are approximately 64.14 and 70.26.
d. To determine the ages that mark the boundaries for the top 10% and bottom 10% of the distribution, we need to calculate the z-scores for these percentiles.
For the top 10%:
z = 1.28 (look up this value in the Normal Curve Tables)
For the bottom 10%:
z = -1.28 (look up this value in the Normal Curve Tables)
Using the z-score formula again, we can find the ages corresponding to these z-scores:
For z = 1.28:
X = M + (z * SD)
X = 67.2 + (1.28 * 3.06)
X ≈ 71.61
For z = -1.28:
X = M + (z * SD)
X = 67.2 + (-1.28 * 3.06)
X ≈ 62.79
Therefore, the ages that mark the boundaries for the top 10% and bottom 10% of the distribution are approximately 62.79 and 71.61, respectively.