np.std() and np.var() for spread in NumPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the standard deviation of the array.

NumPy

import numpy as np
arr = np.array([1, 2, 3, 4, 5])
std_dev = np.[1](arr)
print(std_dev)

Drag options to blanks, or click blank then click option'

Amean

Bstd

Cvar

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.mean() instead of np.std()

Using np.var() which calculates variance, not standard deviation

The np.std() function calculates the standard deviation, which measures how spread out the numbers are.

2fill in blank

medium

Complete the code to calculate the variance of the array.

NumPy

import numpy as np
arr = np.array([10, 20, 30, 40, 50])
variance = np.[1](arr)
print(variance)

Drag options to blanks, or click blank then click option'

Avar

Bstd

Cmean

Dmedian

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.std() instead of np.var()

Using np.mean() which calculates the average, not variance

The np.var() function calculates the variance, which is the average of the squared differences from the mean.

3fill in blank

hard

Fix the error in the code to correctly calculate the standard deviation.

NumPy

import numpy as np
values = [2, 4, 6, 8]
result = np.std([1])
print(result)

Drag options to blanks, or click blank then click option'

Avalues.tolist()

Bvalues

Clist(values)

Dnp.array(values)

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a plain Python list directly without conversion

Using list() or tolist() which do not convert to NumPy array

NumPy functions work best with NumPy arrays. Converting the list to an array ensures correct calculation.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths for words longer than 3 characters.

NumPy

words = ['apple', 'bat', 'carrot', 'dog', 'elephant']
lengths = {word: [1] for word in words if len(word) [2] 3}
print(lengths)

Drag options to blanks, or click blank then click option'

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself instead of its length

Using less than < instead of greater than > for filtering

The dictionary comprehension uses len(word) to get the length and filters words with length greater than 3 using >.

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase words and their lengths for words longer than 4 characters.

NumPy

words = ['tree', 'house', 'river', 'sky', 'mountain']
result = { [1]: [2] for word in words if len(word) [3] 4 }
print(result)

Drag options to blanks, or click blank then click option'

Aword.upper()

Blen(word)

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original word instead of uppercase

Using less than < instead of greater than > for filtering

Using the word itself as value instead of length

The dictionary comprehension uses the uppercase word as the key (word.upper()), the length as the value (len(word)), and filters words longer than 4 characters (>).

Practice

1. What does np.std() measure in a dataset?

easy

A. The maximum value in the dataset

B. The sum of all data points

C. The average distance of data points from the mean in original units

D. The total number of data points

Solution

1. Step 1: Understand the purpose of np.std()

   np.std() calculates the standard deviation, which shows how spread out the data is around the mean in the same units as the data.

2. Step 2: Compare options to definition

   Only The average distance of data points from the mean in original units correctly describes this as the average distance from the mean in original units.

3. Final Answer:

   The average distance of data points from the mean in original units -> Option C

4. Quick Check:

   Standard deviation = average spread in original units [OK]

Hint: Standard deviation = spread in original units [OK]

Common Mistakes:

• Confusing standard deviation with variance
• Thinking it counts data points
• Mixing it with sum or max values

2. Which of the following is the correct syntax to calculate variance of array arr using NumPy?

easy

A. np.std(arr)

B. np.variance(arr)

C. arr.var()

D. np.var(arr)

Solution

1. Step 1: Recall the NumPy function for variance

   The correct NumPy function to calculate variance is np.var().

2. Step 2: Check each option

   np.var(arr) uses np.var(arr) which is correct. np.variance(arr) uses a non-existent function np.variance(). arr.var() uses arr.var() which works only if arr is a NumPy array, but the question asks for NumPy function syntax. np.std(arr) calculates standard deviation, not variance.

3. Final Answer:

   np.var(arr) -> Option D

4. Quick Check:

   Variance function = np.var() [OK]

Hint: Variance uses np.var(), not np.variance() [OK]

Common Mistakes:

• Using np.variance() which does not exist
• Confusing variance with standard deviation function
• Using method on list instead of NumPy array

3. What is the output of the following code?

import numpy as np
arr = np.array([2, 4, 4, 4, 5, 5, 7, 9])
print(np.var(arr))

medium

A. 2.0

B. 4.0

C. 2.5

D. 3.0

Solution

1. Step 1: Calculate the mean of the array

   The array is [2,4,4,4,5,5,7,9]. Sum = 40, count = 8, mean = 40/8 = 5.

2. Step 2: Calculate variance

   Variance = average of squared differences from mean:
   (2-5)^2=9, (4-5)^2=1, (4-5)^2=1, (4-5)^2=1, (5-5)^2=0, (5-5)^2=0, (7-5)^2=4, (9-5)^2=16
   Sum = 9+1+1+1+0+0+4+16 = 32
   Variance = 32/8 = 4.0

3. Step 3: Check code output

   NumPy's np.var() by default calculates population variance (divides by N). So output is 4.0.

4. Final Answer:

   4.0 -> Option B

5. Quick Check:

   Variance = 4.0 [OK]

Hint: Variance = mean of squared differences [OK]

Common Mistakes:

• Calculating sample variance (dividing by N-1) instead of population
• Mixing variance with standard deviation
• Incorrect mean calculation

4. The code below is intended to print the standard deviation of data. What is wrong?

import numpy as np
data = [1, 2, 3, 4, 5]
print(np.std(data, ddof=1))

medium

A. Nothing is wrong; code runs correctly

B. ddof is not a valid argument for np.std()

C. The code will raise a TypeError because data is a list, not an array

D. The code calculates variance, not standard deviation

Solution

1. Step 1: Check if ddof is valid for np.std()

   The ddof parameter (delta degrees of freedom) is valid for np.std() and adjusts divisor for sample standard deviation.

2. Step 2: Check if input type is acceptable

   np.std() accepts lists and converts them internally to arrays, so no error occurs.

3. Final Answer:

   Nothing is wrong; code runs correctly -> Option A

4. Quick Check:

   ddof works with np.std() and lists accepted [OK]

Hint: ddof works with np.std() for sample std [OK]

Common Mistakes:

• Thinking ddof is invalid for standard deviation
• Assuming lists cause errors in NumPy functions
• Confusing variance and standard deviation

5. You have two datasets:
data1 = np.array([10, 12, 14, 16, 18])
data2 = np.array([10, 10, 10, 10, 10])
Which statement is true about their standard deviation and variance?

hard

A. data1 has higher variance and standard deviation than data2

B. data2 has zero variance and standard deviation

C. data2 has higher variance but lower standard deviation than data1

D. Both datasets have the same variance and standard deviation

Solution

1. Step 1: Analyze data2 values

   All values in data2 are the same (10), so variance and standard deviation must be zero because there is no spread.

2. Step 2: Analyze data1 values

   data1 values vary from 10 to 18, so variance and standard deviation are greater than zero, meaning data1 has higher spread.

3. Final Answer:

   data1 has higher variance and standard deviation than data2 -> Option A

4. Quick Check:

   Constant data variance = 0; varying data variance > 0 [OK]

Hint: Constant data has zero variance and std [OK]

Common Mistakes:

• Assuming equal variance for different spread data
• Confusing variance with mean
• Thinking zero variance means error