Summary statistics in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of summary statistics with NA values

What is the output of the following R code that calculates summary statistics on a vector with missing values?

R Programming

x <- c(5, 7, NA, 3, 9, NA, 4)
summary(x)

A Min. : 3.0 1st Qu.: 4.0 Median : 5.0 Mean : 5.6 3rd Qu.: 7.0 Max. : 9.0 NA's : 2

B Min. : 3.0 1st Qu.: 4.0 Median : 5.0 Mean : NA 3rd Qu.: 7.0 Max. : 9.0

C Min. : 3.0 1st Qu.: 4.0 Median : 5.0 Mean : 5.6 3rd Qu.: 7.0 Max. : 9.0

D Min. : NA 1st Qu.: NA Median : NA Mean : NA 3rd Qu.: NA Max. : NA

💡 Hint

Remember that summary() shows count of NA values separately.

The summary() function in R calculates statistics ignoring NA values for numeric vectors but reports the count of NA values separately.

❓ data_output

intermediate

Number of rows after filtering by summary statistics

Given a data frame df with a numeric column 'score', how many rows remain after filtering out rows where 'score' is below the median?

R Programming

df <- data.frame(score = c(10, 20, 15, 30, 25))
median_score <- median(df$score)
df_filtered <- df[df$score >= median_score, ]
nrow(df_filtered)

A 3

B 2

C 4

D 5

💡 Hint

Calculate the median and count how many values are equal or above it.

Median is 20; values >= 20 are 20, 30, 25, so 3 rows remain.

❓ visualization

advanced

Interpreting boxplot summary statistics

Which statement correctly describes the boxplot summary statistics shown for a numeric vector in R?

R Programming

x <- c(2, 4, 5, 6, 8, 9, 10, 11, 12)
boxplot(x)

A The median is 10, and the IQR is 6.

B The median is 8, and the interquartile range (IQR) is 10.

C The median is 8, and the IQR is 6.

D The median is 10, and the IQR is 10.

💡 Hint

Recall median is the middle value; IQR is Q3 minus Q1.

Median is the middle value 8; Q1 is 5, Q3 is 11, so IQR = 11 - 5 = 6.

🧠 Conceptual

advanced

Understanding skewness from summary statistics

If a numeric dataset has mean > median > mode, what can you infer about its skewness?

A The distribution is negatively skewed (left-skewed).

B The distribution is positively skewed (right-skewed).

C The distribution is symmetric.

D The distribution is bimodal.

💡 Hint

Think about how mean, median, and mode relate in skewed data.

When mean > median > mode, the tail is longer on the right side, indicating positive skew.

🔧 Debug

expert

Identify the error in summary statistics calculation

What error will this R code produce when calculating the mean of a data frame column?

R Programming

df <- data.frame(a = c(1, 2, 3), b = c('x', 'y', 'z'))
mean(df)

A No error; returns mean of column 'a'

B Error in mean.default(df) : 'trim' must be numeric of length one

C Error in mean.default(df) : argument is not numeric or logical: returning NA

D Error in mean.default(df) : 'x' must be numeric

💡 Hint

mean() expects a numeric vector, not a data frame.

mean() cannot operate on a data frame directly; it requires a numeric vector.

Practice

1. What does the summary() function in R provide when applied to a numeric vector?

easy

A. Minimum, 1st quartile, median, mean, 3rd quartile, and maximum values

B. Only the mean and median values

C. The sum and product of all elements

D. The count of unique values

Solution

1. Step 1: Understand what summary() does for numeric vectors

   The summary() function returns key statistics including minimum, quartiles, median, mean, and maximum.

2. Step 2: Compare options with the known output

   Minimum, 1st quartile, median, mean, 3rd quartile, and maximum values lists all these statistics correctly, while others miss important parts or give unrelated info.

3. Final Answer:

   Minimum, 1st quartile, median, mean, 3rd quartile, and maximum values -> Option A

4. Quick Check:

   summary() = min, Q1, median, mean, Q3, max [OK]

Hint: Remember summary() shows min, quartiles, median, mean, max [OK]

Common Mistakes:

• Thinking summary() only gives mean and median
• Confusing summary() with sum() or length()
• Expecting summary() to count unique values

2. Which of the following is the correct way to get summary statistics of a data frame named df in R?

easy

A. stats(df)

B. summarize(df)

C. summary(df)

D. summary_stats(df)

Solution

1. Step 1: Recall the correct function name for summary statistics in R

   The built-in function to get summary statistics is summary().

2. Step 2: Check the options for correct syntax

   Only summary(df) uses the correct function summary() applied to the data frame df.

3. Final Answer:

   summary(df) -> Option C

4. Quick Check:

   summary() is the correct function [OK]

Hint: Use summary() for data frames, not summarize() or others [OK]

Common Mistakes:

• Using summarize() which is from dplyr, not base R
• Typing summary_stats() which doesn't exist
• Using stats() which is unrelated

3. What will be the output of the following R code?

vec <- c(5, 10, 15, 20, 25)
summary(vec)

medium

A. Min. 5, 1st Qu. 10, Median 15, Mean 15, 3rd Qu. 20, Max. 25

B. Min. 5, Median 10, Mean 15, Max. 25

C. Min. 5, Max. 25, Sum 75

D. Error: summary() cannot be applied to vectors

Solution

1. Step 1: Calculate summary statistics for the vector

   The vector is 5, 10, 15, 20, 25. Minimum is 5, 1st quartile is 10, median is 15, mean is (5+10+15+20+25)/5 = 15, 3rd quartile is 20, maximum is 25.

2. Step 2: Match calculated values with options

   Min. 5, 1st Qu. 10, Median 15, Mean 15, 3rd Qu. 20, Max. 25 matches all these values exactly. Other options miss quartiles or give errors.

3. Final Answer:

   Min. 5, 1st Qu. 10, Median 15, Mean 15, 3rd Qu. 20, Max. 25 -> Option A

4. Quick Check:

   summary(vec) = min 5, Q1 10, median 15, mean 15, Q3 20, max 25 [OK]

Hint: Calculate min, quartiles, median, mean, max for vector [OK]

Common Mistakes:

• Forgetting to calculate quartiles
• Confusing median with mean
• Expecting an error for summary() on vectors

4. Identify the error in this R code that tries to get summary statistics of a numeric vector nums:

summary(nums, digits=2)

medium

A. The vector nums must be sorted before summary()

B. The vector nums must be converted to a data frame first

C. The function should be summarize() instead of summary()

D. The summary() function does not have a digits argument

Solution

1. Step 1: Check the arguments of summary() in R

   The base R summary() function does not accept a digits argument. This causes an error.

2. Step 2: Verify other options

   Converting to data frame or sorting is not required. summarize() is from dplyr, not base R.

3. Final Answer:

   The summary() function does not have a digits argument -> Option D

4. Quick Check:

   summary() has no digits argument [OK]

Hint: Check function arguments carefully to avoid errors [OK]

Common Mistakes:

• Assuming summary() accepts digits argument
• Thinking vector must be data frame
• Confusing summary() with dplyr's summarize()

5. You have a data frame df with a numeric column age. You want to find the median age but exclude any missing values (NA). Which R code correctly does this?

hard

A. median(df$age)

B. median(df$age, na.rm = TRUE)

C. summary(df$age, na.rm = TRUE)["Median"]

D. summary(df$age)$median

Solution

1. Step 1: Understand how to calculate median excluding NA values

   The median() function has an argument na.rm = TRUE to remove missing values before calculation.

2. Step 2: Check each option for correctness

   median(df$age, na.rm = TRUE) correctly uses median() with na.rm = TRUE. summary(df$age, na.rm = TRUE)["Median"] tries to extract median from summary but summary() does not accept na.rm. median(df$age) ignores NA values causing NA result. summary(df$age)$median tries to access median as a list element which is invalid.

3. Final Answer:

   median(df$age, na.rm = TRUE) -> Option B

4. Quick Check:

   Use median() with na.rm=TRUE to exclude NA [OK]

Hint: Use median(..., na.rm=TRUE) to ignore missing values [OK]

Common Mistakes:

• Not using na.rm=TRUE causing NA result
• Trying to get median from summary() output incorrectly
• Assuming summary() removes NA automatically