Random sampling distributions in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of sampling from a normal distribution

What is the shape of the array produced by this code?

NumPy

import numpy as np
sample = np.random.normal(loc=0, scale=1, size=(5, 3))
print(sample.shape)

A (5, 3)

B (3, 5)

C (5,)

D (15,)

💡 Hint

The size parameter defines the shape of the output array.

The size=(5,3) argument creates a 2D array with 5 rows and 3 columns.

❓ data_output

intermediate

Mean of samples from uniform distribution

What is the approximate mean of the samples generated by this code?

NumPy

import numpy as np
samples = np.random.uniform(low=10, high=20, size=10000)
print(round(samples.mean(), 1))

A 15.0

B 10.0

C 20.0

D 5.0

💡 Hint

The mean of a uniform distribution is the midpoint between low and high.

The mean of uniform(10, 20) is (10 + 20) / 2 = 15.

❓ visualization

advanced

Histogram shape of binomial samples

Which histogram shape best describes the samples from this binomial distribution?

NumPy

import numpy as np
import matplotlib.pyplot as plt
samples = np.random.binomial(n=10, p=0.5, size=1000)
plt.hist(samples, bins=range(12), edgecolor='black')
plt.show()

A Single spike at 10

B Symmetric bell-shaped around 5

C Skewed heavily to the right

D Uniform flat distribution

💡 Hint

Binomial with p=0.5 tends to be symmetric around n*p.

The binomial distribution with n=10 and p=0.5 is symmetric and peaks near 5.

🧠 Conceptual

advanced

Effect of sample size on sample mean variance

If you increase the sample size from 10 to 1000 when sampling from a normal distribution, what happens to the variance of the sample mean?

A It stays the same because variance depends only on the distribution

B It increases because more data adds more variability

C It decreases because larger samples give more stable means

D It becomes zero because large samples have no variance

💡 Hint

Think about how averaging many values affects variability.

The variance of the sample mean decreases as sample size increases, making the mean more stable.

🔧 Debug

expert

Identify the error in random sampling code

What error does this code raise when run?

NumPy

import numpy as np
samples = np.random.normal(loc=0, scale=1, size=-5)
print(samples)

A No error, prints an empty array

B TypeError: size must be an integer or tuple of integers

C SyntaxError: invalid syntax

D ValueError: negative dimensions are not allowed

💡 Hint

Check the size parameter for validity.

Negative size is invalid and raises a ValueError in numpy random functions.

Practice

1. What does the numpy.random.choice function do in random sampling?

easy

A. It selects random elements from a given array or list.

B. It generates random numbers from a normal distribution.

C. It creates a sequence of numbers with equal spacing.

D. It calculates the mean of a dataset.

Solution

1. Step 1: Understand the purpose of numpy.random.choice

   This function is used to randomly pick elements from an input array or list, optionally with replacement.

2. Step 2: Compare with other options

   Options A, B, and C describe different numpy functions or operations unrelated to random selection from an array.

3. Final Answer:

   It selects random elements from a given array or list. -> Option A

4. Quick Check:

   numpy.random.choice = random selection [OK]

Hint: Remember: choice picks from arrays, normal generates bell curve numbers [OK]

Common Mistakes:

• Confusing choice with normal distribution generator
• Thinking choice creates sequences
• Assuming choice calculates statistics

2. Which of the following is the correct syntax to generate 5 samples from a binomial distribution with 10 trials and success probability 0.3 using numpy?

easy

A. numpy.binomial(10, 0.3, 5)

B. numpy.random.binomial(10, 0.3, size=5)

C. numpy.random.binomial(size=5, 10, 0.3)

D. numpy.random.binomial(5, 10, 0.3)

Solution

1. Step 1: Recall the numpy binomial function signature

   The correct syntax is numpy.random.binomial(n, p, size) where n is number of trials, p is success probability, and size is number of samples.

2. Step 2: Check each option

   numpy.random.binomial(10, 0.3, size=5) matches the correct order and uses the keyword size. numpy.binomial(10, 0.3, 5) misses .random. numpy.random.binomial(size=5, 10, 0.3) mixes keyword size=5 followed by positionals, causing SyntaxError. numpy.random.binomial(5, 10, 0.3) swaps parameters incorrectly.

3. Final Answer:

   numpy.random.binomial(10, 0.3, size=5) -> Option B

4. Quick Check:

   binomial(n, p, size) = numpy.random.binomial(10, 0.3, size=5) [OK]

Hint: Remember order: trials, probability, then size keyword [OK]

Common Mistakes:

• Mixing argument order
• Omitting size keyword
• Using wrong function name

3. What is the output of the following code?

import numpy as np
np.random.seed(0)
samples = np.random.normal(loc=5, scale=2, size=3)
print(np.round(samples, 2))

medium

A. [7.65 5.12 6.34]

B. [8.53 4.32 6.58]

C. [8.82 4.0 6.89]

D. [8.53 5.8 6.96]

Solution

1. Step 1: Understand the code

   The code sets a random seed for reproducibility, then generates 3 samples from a normal distribution with mean 5 and standard deviation 2.

2. Step 2: Run or recall the output

   Using seed 0, the samples generated are [8.53, 5.8, 6.96] after rounding to 2 decimals.

3. Final Answer:

   [8.53 5.8 6.96] -> Option D

4. Quick Check:

   Seed 0 normal samples = [8.53 5.8 6.96] [OK]

Hint: Use seed for fixed output, then round to compare [OK]

Common Mistakes:

• Ignoring seed causes different outputs
• Rounding incorrectly
• Mixing mean and scale parameters

4. Identify the error in this code snippet that tries to sample 4 values from a uniform distribution between 0 and 1:

import numpy as np
samples = np.random.uniform(low=0, high=1, size=4)
print(samples)

medium

A. There is no error; the code runs correctly.

B. The parameters low and high are reversed.

C. The function name should be random.uniform without numpy prefix.

D. The size parameter must be a tuple, not an integer.

Solution

1. Step 1: Review the numpy uniform function syntax

   np.random.uniform(low, high, size) is the correct way to generate samples from a uniform distribution.

2. Step 2: Check the code correctness

   The code uses correct function name, parameter order, and size as integer for 1D array. It will run without error and print 4 random values between 0 and 1.

3. Final Answer:

   There is no error; the code runs correctly. -> Option A

4. Quick Check:

   Uniform sampling syntax correct = There is no error; the code runs correctly. [OK]

Hint: Check function name and parameter order carefully [OK]

Common Mistakes:

• Confusing numpy.random.uniform with Python's random.uniform
• Swapping low and high values
• Using tuple for size unnecessarily

5. You want to simulate rolling a fair six-sided die 1000 times and count how many times the number 4 appears. Which numpy code correctly performs this simulation?

hard

A. rolls = np.random.choice([1,2,3,4,5,6], size=1000)
count_4 = np.sum(rolls == 4)

B. rolls = np.random.binomial(6, 1/6, size=1000)
count_4 = np.count_nonzero(rolls == 4)

C. rolls = np.random.randint(1, 7, size=1000)
count_4 = np.sum(rolls == 4)

D. rolls = np.random.normal(4, 1, size=1000)
count_4 = np.sum(rolls == 4)

Solution

1. Step 1: Understand the problem

   We want to simulate rolling a die 1000 times, each roll equally likely from 1 to 6, then count how many times 4 appears.

2. Step 2: Evaluate each option

   rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4): rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) uses randint from 1 to 6 (exclusive upper bound 7), perfect for die rolls. rolls = np.random.binomial(6, 1/6, size=1000) count_4 = np.count_nonzero(rolls == 4): rolls = np.random.binomial(6, 1/6, size=1000) count_4 = np.count_nonzero(rolls == 4) uses binomial which models number of successes in trials, not die rolls. rolls = np.random.choice([1,2,3,4,5,6], size=1000) count_4 = np.sum(rolls == 4): rolls = np.random.choice([1,2,3,4,5,6], size=1000) count_4 = np.sum(rolls == 4) omits replace=True (defaults to False), raising ValueError since size=1000 > 6. rolls = np.random.normal(4, 1, size=1000) count_4 = np.sum(rolls == 4): rolls = np.random.normal(4, 1, size=1000) count_4 = np.sum(rolls == 4) uses normal distribution, inappropriate for discrete die rolls.

3. Step 3: Choose the best correct code

   rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) is the correct way to simulate die rolls and count 4s.

4. Final Answer:

   rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) -> Option C

5. Quick Check:

   Use randint for discrete uniform die rolls = rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) [OK]

Hint: Use randint(1,7) for dice rolls, then count matches [OK]

Common Mistakes:

• Using binomial instead of uniform discrete
• Using normal distribution for discrete outcomes
• Forgetting replace=True in choice