Assume that you run the following code. just_above_ave = min(student$grade[which(student$grade = mean(student$grade))]) Then, you decided to get the same result in an alternative way. df5 % filter(grade = mean(grade)) %% summarise(min(grade)) the value in df5[1,1] is equal to just_above_ave.

Question

Assume that you run the following code.

just_above_ave = min(student$grade[which(student$grade >= mean(student$grade))])

Then, you decided to get the same result in an alternative way.

df5 <- student %>% 
  filter(grade >= mean(grade)) %>% 
  summarise(min(grade))

the value in df5[1,1] is equal to just_above_ave.

BlackTom AI · Accepted Answer

Let's examine what each piece of code does step by step, and then compare the results.

Option assessment start with the first approach: just_above_ave = min(student$grade[which(student$grade >= mean(student$grade))]).
- This computes the mean of all grades: mean(student$grade).
- It then selects all grades that are greater than or equal to that mean: student$grade[student$grade >= mean(student$grade)].
- It applies min(...) to these selected grades, giving the smallest grade that is still at least the mean. If there are any ties at the boundary, the minimum among them is returned.

Now consider the dplyr pipeline: df5 <- student %>% filter(grade >= mean(grade)) %>% summarise(min(grade)).
- The mean(grade) inside filter uses the same data frame (assuming no grouping and no special scoping effects), so it should compute the same threshold as mean(student$grade) above, provided there are no NA complications.
- The filter step keeps rows where grade is greater than or equal to that threshold, just like the base R expression.
- The summarise(min(grade)) then computes the minimum grade among the remaining rows, i.e., among those with grade >= mean(grade).

Key subtleties that could affect equality:
- NA handling: If there are missing values in student$grade, mean(student$grade) will be NA unless na.rm = TRUE is specified. In that case, the base R expression would yield NA, and the dplyr version would also yield NA, potentially producing different results or an error depending on how NA is propagated. If there are no NAs (or na.rm is consistently applied in both contexts), both expressions operate on the same set of rows.
- Environment and scoping: In standard usage with a single data frame and no grouping, mean(grade) inside the dplyr chain behaves like mean(student$grade). If there were groups or summarise semantics different from the base R approach, results could diverge, but those conditions are not indicated here.
- Boundary behavior: Both approaches include the boundary grade values (those equal to the mean), so the minimum among those should coincide in the usual case.

Given typical assumptions (no NA issues, no grouping, identical definitions of mean in both expressions), both approaches identify the same subset of rows (grade >= mean(grade)) and then take the minimum grade among that subset. Therefore, the value stored in df5[1,1] should equal just_above_ave.

If you encounter NA values and do not handle na.rm consistently (e.g., mean(..., na.rm = TRUE) in both expressions), you would not get a reliable equality, and the statement would be false in that scenario. In a clean dataset with no NAs, the statement is true.

COMM_V 205 101 102 103 2025W1 R Review Quiz

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