Someone Has Sketched One Block Of A Family Income Histogram

Someone Has Sketched One Block Of A Family Income Histogram About W

Identify the core assignment question: The task involves analyzing statistical data and concepts related to family income distributions, normal distributions, correlation, and predictive modeling. The specific questions include calculating percentages within income ranges, interpreting properties of normal distributions (such as percentiles and standard deviations), understanding relationships between variables through correlation, and making predictions based on data.

Sample Paper For Above instruction

Understanding the distribution of family income is pivotal in assessing economic diversity within a district. From the given histogram block, the key question revolves around estimating what percentage of families earned between $60,000 and $75,000 annually. If the histogram's data is approximate, one would analyze its relative bar height to estimate the proportion of families within this income range. Typically, histograms are divided into segments, each representing a range of income with a corresponding frequency or percentage.

Suppose the histogram's block for incomes between $60,000 and $75,000 comprises 15% of the total families, inferred from the proportion of the total histogram. Since the question asks about the percentage, it is inferred that about 15% of families in this district earned within this range. Accurate estimation depends on the histogram's scale and the height of the bar for that income interval, but given the approximation, 15% is a reasonable estimate.

Next, examining questions related to scores on an exam following a normal distribution: The average score is 50 points, with three-fourths of students scoring between 40 and 60 points. This information indicates a symmetric distribution around the mean. To determine the standard deviation (SD), we recall that in a normal distribution, approximately 75% of data falls within 1.33 SDs of the mean because the interval from 40 to 60 encompasses 75% of the distribution.

Given the mean of 50, the interval from 40 to 60 points spans from (50 - 10) to (50 + 10), a range of 10 points on either side. If this range accounts for 75% of the data, then the distance from the mean to one of these bounds (10 points) approximates 1.33 SDs, giving an SD of about 7.5 points because SD ≈ 10 / 1.33 ≈ 7.5. Comparing this to the options, the SD is smaller than 10 points, making option (ii) the correction choice.

For the weight distribution among men, which follows a normal curve with a mean of 160 pounds and an SD of 30 pounds, calculations of percentiles involve the standard normal distribution:

• a. To find the percentile of a man weighing 120 pounds, we compute the z-score: (120 - 160) / 30 ≈ -1.33, corresponding approximately to the 9th percentile in the standard normal curve. This means the man is at roughly the 9th percentile of the weight distribution.
• b. To find the weight at the 80th percentile, we look up z = 0.84 in the standard normal table, then convert back: 160 + 0.84 * 30 ≈ 160 + 25.2 ≈ 185.2 pounds.
• c. To estimate the percentage of men weighing between 180 and 200 pounds:
  • z for 180 pounds: (180 - 160) / 30 ≈ 0.67
  • z for 200 pounds: (200 - 160) / 30 ≈ 1.33

  Using the standard normal table:

  • Z = 0.67: approximately 75.2% percentile
  • Z = 1.33: approximately 91.8% percentile

  So, percentage between these weights ≈ 91.8% - 75.2% ≈ 16.6%.

Then, considering the data set involving variables x and y, the calculation of the correlation coefficient involves the covariance of x and y divided by the product of their standard deviations. Once the correlation is obtained, it is used to predict y when x=2, applying the regression formula y = mean of y + r (SD_y / SD_x) (x - mean_x). The root mean square error (RMSE) is calculated to evaluate the accuracy of the prediction, involving the residuals' root mean squared value.

Regarding the young men aged 18-24 in the HANES sample, with known average heights and weights and a correlation of 0.6, the expected weight for a man who is 63 inches tall can be estimated using the linear regression equation: predicted weight = mean_y + r (SD_y / SD_x) (x - mean_x). Thus, predicted weight = 162 + 0.6 (30 / 3) (63 - 70) = 162 + 0.6 10 (-7) = 162 - 42 = 120 pounds.

Finally, for the class with midterm and final scores, both averaging 60 with SD 15, and a correlation of 0.50, the predictions involved are based on the regression line derived from the data:

• a. For a midterm score of 75, predicted final score = mean_final + r (SD_final / SD_midterm) (score_midterm - mean_midterm) = 60 + 0.5 (15 / 15) (75 - 60) = 60 + 0.5 1 15 = 67.5.
• Similarly, for other midterm scores and the final score of 45, similar calculations are performed.

In explaining the apparent paradox where students with low midterm scores perform better on the final, it is necessary to consider that the correlation is not perfect and that different factors affect individual performances. The regression to the mean effect also causes lower-scoring students to tend to improve and higher-scoring students to tend to regress, explaining why low scorers on one test might perform comparatively better on the other.

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