Descriptive statistics (mean, median, std) in MATLAB - Time & Space Complexity

Time Complexity: Descriptive statistics (mean, median, std)

O(n)

Understanding Time Complexity

We want to understand how the time to calculate mean, median, and standard deviation changes as the data size grows.

How does the work increase when we have more numbers to analyze?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

data = rand(1, n);  % Generate n random numbers
mean_val = mean(data);  % Calculate mean
median_val = median(data);  % Calculate median
std_val = std(data);  % Calculate standard deviation

This code generates a list of n numbers and calculates three common descriptive statistics.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Each function (mean, median, std) processes the entire data array.
• How many times: Each function goes through the n numbers once or a few times internally.

How Execution Grows With Input

As the number of data points n grows, the time to calculate mean, median, and std grows roughly in direct proportion.

Input Size (n)	Approx. Operations
10	About 30 (3 functions x 10 elements each)
100	About 300
1000	About 3000

Pattern observation: The work grows linearly as the data size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to calculate these statistics grows in a straight line with the number of data points.

Common Mistake

[X] Wrong: "Calculating median is much slower than mean because it sorts the data every time."

[OK] Correct: While median often uses sorting internally, efficient algorithms or MATLAB's built-in functions handle this quickly, and the overall time still grows linearly with data size.

Interview Connect

Knowing how descriptive statistics scale helps you understand data processing limits and optimize analysis steps in real projects.

Self-Check

"What if we calculate these statistics repeatedly inside a loop for m different datasets each of size n? How would the time complexity change?"

Practice

1. What does the mean function calculate in MATLAB when applied to a numeric array?

easy

A. The difference between the largest and smallest values

B. The middle value after sorting the array

C. The total sum of all elements in the array

D. The average value of all elements in the array

Solution

1. Step 1: Understand the mean function

   The mean function calculates the average by adding all elements and dividing by the number of elements.

2. Step 2: Compare with other statistics

   Median is the middle value, not the average. Sum is total, not average. Range is difference, not average.

3. Final Answer:

   The average value of all elements in the array -> Option D

4. Quick Check:

   Mean = Average [OK]

Hint: Mean is the sum divided by count [OK]

Common Mistakes:

• Confusing mean with median
• Thinking mean is the sum only
• Mixing mean with range

2. Which of the following is the correct MATLAB syntax to calculate the standard deviation of a vector data?

easy

A. median(data)

B. standard_deviation(data)

C. std(data)

D. mean(data)

Solution

1. Step 1: Recall MATLAB function names

   MATLAB uses std() to calculate standard deviation, not standard_deviation().

2. Step 2: Eliminate incorrect options

   mean() calculates average, median() calculates middle value, so both are incorrect for standard deviation.

3. Final Answer:

   std(data) -> Option C

4. Quick Check:

   Standard deviation function is std() [OK]

Hint: Use std() for standard deviation in MATLAB [OK]

Common Mistakes:

• Using mean() instead of std()
• Typing non-existent function standard_deviation()
• Confusing median() with std()

3. What is the output of the following MATLAB code?

data = [3, 7, 5, 9, 1];
m = median(data);
s = std(data);
[m, s]

medium

A. [5, 2.8284]

B. [5, 3.1623]

C. [7, 3.1623]

D. [3, 2.8284]

Solution

1. Step 1: Calculate the median of data

   Sorted data is [1, 3, 5, 7, 9]. The middle value is 5, so median = 5.

2. Step 2: Calculate the standard deviation

   Mean = (3+7+5+9+1)/5 = 5. Differences squared: (3-5)^2=4, (7-5)^2=4, (5-5)^2=0, (9-5)^2=16, (1-5)^2=16. Sum = 40, variance = 40/4 = 10 (using n-1 for std), std = sqrt(10) ≈ 3.1623.

3. Final Answer:

   [5, 3.1623] -> Option B

4. Quick Check:

   Median = 5, Std ≈ 3.16 [OK]

Hint: Median is middle; std uses n-1 in denominator [OK]

Common Mistakes:

• Using n instead of n-1 for std denominator
• Mixing mean and median values
• Incorrectly sorting data for median

4. Identify the error in this MATLAB code snippet that calculates the mean and median of values:

values = [10, 20, 30, 40];
mean_val = mean(values)
median_val = median(values

medium

A. Missing closing parenthesis in median function call

B. Using mean instead of median for median_val

C. Incorrect variable name 'values'

D. No error, code runs fine

Solution

1. Step 1: Check syntax of median function call

   The line median_val = median(values is missing a closing parenthesis.

2. Step 2: Confirm other parts are correct

   mean(values) is correct, variable name is consistent, so no other errors.

3. Final Answer:

   Missing closing parenthesis in median function call -> Option A

4. Quick Check:

   Syntax error due to missing ')' [OK]

Hint: Check all parentheses in function calls [OK]

Common Mistakes:

• Ignoring missing parenthesis errors
• Confusing mean and median functions
• Assuming variable name errors without checking

5. Given a dataset data = [2, 4, 4, 4, 5, 5, 7, 9], which MATLAB code correctly calculates the mean, median, and standard deviation and stores them in variables m, med, s respectively?

hard

A. m = mean(data); med = median(data); s = std(data);

B. m = median(data); med = mean(data); s = std(data);

C. m = std(data); med = mean(data); s = median(data);

D. m = mean(data); med = std(data); s = median(data);

Solution

1. Step 1: Match variables to correct functions

   Variable m should hold mean, med median, s standard deviation.

2. Step 2: Verify correct function usage

   m = mean(data); med = median(data); s = std(data); assigns mean to m, median to med, std to s, which is correct.

3. Final Answer:

   m = mean(data); med = median(data); s = std(data); -> Option A

4. Quick Check:

   Correct variable-function mapping [OK]

Hint: Assign mean, median, std to matching variable names [OK]

Common Mistakes:

• Swapping mean and median assignments
• Assigning std to median variable
• Mixing variable names and functions