Problem Solving and Data Analysis Flashcards

A researcher collects data on monthly subscription costs (in dollars) for 12 streaming services: 8, 10, 10, 12, 13, 14, 15, 15, 15, 18, 20, 120. Which measure of center would be MOST misleading when describing the typical cost of these services, and why?

Answer: The mean, because the outlier of $120 disproportionately inflates it

The mean is (8+10+10+12+13+14+15+15+15+18+20+120)/12 = 270/12 = $22.50, which is higher than 11 of the 12 values. The single outlier ($120) pulls the mean far above what most services cost, making it misleading. The median (14.5) and mode (15) better represent the typical cost. Range is not a measure of center at all.

In a study of exam scores, the mean score is 74 with a standard deviation of 8. A student scored at the 84th percentile. If the distribution is approximately normal, which score is closest to the student's score?

Answer: 82

In a normal distribution, the 84th percentile corresponds approximately to one standard deviation above the mean (z ≈ 1.0). The score is therefore 74 + 1(8) = 82. The 84th percentile is a well-known landmark: roughly 84% of values fall below z = +1. Choices 84, 88, and 90 would correspond to higher z-scores and higher percentiles.

A scatterplot shows the relationship between hours studied per week (x) and GPA (y) for 80 college students. The correlation coefficient is r = 0.72, and the least-squares regression line is ŷ = 1.8 + 0.15x. A student who studies 20 hours per week is predicted to have a GPA of 4.8. What is the most appropriate conclusion?

Answer: The prediction of 4.8 is an example of extrapolation beyond the data's range, since GPA maxes at 4.0

Plugging x = 20 into the equation gives ŷ = 1.8 + 0.15(20) = 1.8 + 3.0 = 4.8. Since GPA is bounded at 4.0, this prediction is impossible — it results from extrapolating the linear model beyond the meaningful range of the response variable. The model may be valid within the observed data range but breaks down outside it. This is a classic extrapolation/model limitation issue.

Two surveys measure job satisfaction on a 1–10 scale. Survey A has a mean of 7.2 and a standard deviation of 0.4. Survey B has a mean of 7.2 and a standard deviation of 2.9. Which statement best describes the difference between the two distributions?

Answer: Survey A respondents are more consistently satisfied; Survey B has much greater variability in satisfaction levels

Standard deviation measures spread, not sample size or reliability. Survey A's tiny SD (0.4) means nearly all respondents rated satisfaction close to 7.2 — high consistency. Survey B's large SD (2.9) means ratings are widely spread across the 1–10 scale, indicating very mixed satisfaction. Same mean does not mean same distribution; variability tells the rest of the story. A smaller SD does not imply more outliers — it implies the opposite.

A company tests two manufacturing processes. Process X produces parts with lengths (in mm): 50.1, 50.3, 49.9, 50.2, 50.0. Process Y produces: 48.0, 52.1, 49.5, 51.8, 50.1. Both processes have a mean of approximately 50.1 mm. A part is acceptable if its length is within 1 mm of 50 mm (i.e., between 49 and 51 mm). Which process is preferable, and what statistical concept supports this?

Answer: Process X, because its lower variability (all values within 0.4 mm of the mean) means nearly all parts fall within the acceptable range

Process X values (49.9–50.3) all fall within the 49–51 mm acceptable range. Process Y values include 48.0 and 52.1, which fall outside the acceptable range — meaning defective parts are produced. Even though both processes have the same mean (~50.1), Process X's low variability makes it far superior for quality control. This illustrates why variability, not just center, matters in applied data analysis.

A data set of 200 values has a median of 45 and an interquartile range (IQR) of 18. Values below Q1 − 1.5(IQR) or above Q3 + 1.5(IQR) are considered outliers. If Q1 = 36 and Q3 = 54, which of the following values would NOT be classified as an outlier?

Answer: 28

With Q1 = 36 and IQR = 18: the lower fence = 36 − 1.5(18) = 36 − 27 = 9, and the upper fence = 54 + 1.5(18) = 54 + 27 = 81. Outliers are values below 9 or above 81. Value 8 is below 9 → outlier. Value 28 is between 9 and 81 → NOT an outlier. Value 81 equals the upper fence — values must be strictly greater than 81 to be outliers, so 81 is borderline; however value 83 > 81 → outlier. Among the choices, 28 is clearly not an outlier.