Random variable generation in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of generating normal random variables

What is the output of this code snippet that generates 3 samples from a normal distribution with mean 0 and standard deviation 1 using scipy.stats?

SciPy

from scipy.stats import norm
samples = norm.rvs(loc=0, scale=1, size=3)
samples_rounded = [round(x, 2) for x in samples]
samples_rounded

A [0.0, 0.0, 0.0]

B A list of 3 float numbers close to 0 but random, e.g., [0.12, -0.45, 1.03]

C [0.5, -1.2, 0.3]

D [1, 1, 1]

💡 Hint

The output is random but will be three float numbers around zero.

The norm.rvs function generates random samples from a normal distribution. Since the mean is 0 and std dev is 1, the values will be floats near zero but random each time.

❓ data_output

intermediate

Shape of generated random variable array

What is the shape of the array generated by this code that samples from a uniform distribution?

SciPy

from scipy.stats import uniform
samples = uniform.rvs(loc=0, scale=10, size=(4, 2))
samples.shape

A (4, 2)

B (2, 4)

C (8,)

D (4,)

💡 Hint

The size parameter defines the shape of the output array.

The size=(4, 2) argument tells scipy to generate a 2D array with 4 rows and 2 columns.

🔧 Debug

advanced

Identify the error in random variable generation code

What error does this code raise when trying to generate samples from a binomial distribution?

SciPy

from scipy.stats import binom
samples = binom.rvs(n=10, p=0.5, size='5')

A TypeError: 'str' object cannot be interpreted as an integer

B ValueError: Probability p must be between 0 and 1

C SyntaxError: invalid syntax

D No error, code runs fine

💡 Hint

Check the type of the size argument.

The size parameter must be an integer or tuple of integers, not a string. Passing '5' causes a TypeError.

🚀 Application

advanced

Generating and summarizing Poisson random variables

You generate 1000 samples from a Poisson distribution with lambda=3. Which option correctly describes the expected mean and variance of the samples?

SciPy

from scipy.stats import poisson
samples = poisson.rvs(mu=3, size=1000)
mean_val = samples.mean()
var_val = samples.var()

A Mean close to 0, variance close to 3

B Mean close to 3, variance close to 0

C Mean and variance both close to 3

D Mean and variance both close to 0

💡 Hint

For Poisson distribution, mean equals variance.

Poisson distribution has mean and variance equal to lambda (mu). So both mean and variance should be near 3.

🧠 Conceptual

expert

Effect of random seed on reproducibility

Which statement about setting a random seed before generating random variables with scipy.stats is correct?

A Setting the seed has no effect on random variable generation.

B Setting the seed makes the random numbers more random.

C Setting the seed disables randomness and returns constant values.

D Setting the seed ensures the same random numbers are generated every time the code runs.

💡 Hint

Think about reproducibility in experiments.

Setting a random seed initializes the random number generator to a fixed state, so the same sequence of random numbers is produced each run.

Practice

1. What does the rvs method in scipy.stats do?

easy

A. Calculates the mean of a dataset

B. Sorts an array in ascending order

C. Plots a histogram of data

D. Generates random samples from a specified probability distribution

Solution

1. Step 1: Understand the purpose of rvs

   The rvs method is used to create random values that follow a given distribution.

2. Step 2: Compare with other options

   Calculating mean, plotting, or sorting are different functions and not related to rvs.

3. Final Answer:

   Generates random samples from a specified probability distribution -> Option D

4. Quick Check:

   rvs = random samples [OK]

Hint: Remember: rvs means random variable samples [OK]

Common Mistakes:

• Confusing rvs with statistical calculations like mean
• Thinking rvs plots data
• Assuming rvs sorts data

2. Which of the following is the correct way to generate 5 random samples from a normal distribution with mean 0 and standard deviation 1 using scipy.stats?

easy

A. norm.generate(5, mean=0, sd=1)

B. norm.rvs(loc=0, scale=1, size=5)

C. stats.norm.sample(0, 1, 5)

D. norm.random(mean=0, std=1, count=5)

Solution

1. Step 1: Recall the correct method and parameters

   In scipy.stats.norm, rvs generates samples. Parameters are loc for mean, scale for std deviation, and size for number of samples.

2. Step 2: Check each option

   Only norm.rvs(loc=0, scale=1, size=5) uses correct method rvs and correct parameter names.

3. Final Answer:

   norm.rvs(loc=0, scale=1, size=5) -> Option B

4. Quick Check:

   Correct method and parameters = norm.rvs(loc=0, scale=1, size=5) [OK]

Hint: Use rvs with loc, scale, and size for sampling [OK]

Common Mistakes:

• Using wrong method names like random or sample
• Mixing parameter names like mean instead of loc
• Incorrect order or missing size parameter

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

from scipy.stats import uniform
samples = uniform.rvs(loc=10, scale=5, size=3)
print(type(samples))

medium

A. <class 'numpy.ndarray'>

B. <class 'list'>

C. <class 'tuple'>

D. <class 'int'>

Solution

1. Step 1: Understand output of rvs with size>1

   When size is more than 1, rvs returns a NumPy array of samples.

2. Step 2: Check the printed type

   Printing type(samples) will show <class 'numpy.ndarray'>.

3. Final Answer:

   <class 'numpy.ndarray'> -> Option A

4. Quick Check:

   rvs with size=3 returns ndarray [OK]

Hint: rvs with size>1 returns numpy array [OK]

Common Mistakes:

• Assuming output is a list instead of ndarray
• Thinking output is a single number
• Confusing tuple with array

4. Identify the error in the following code snippet:

from scipy.stats import binom
samples = binom.rvs(n=10, p=0.5, size='5')

medium

A. Parameter 'size' should be an integer, not a string

B. Parameter 'p' must be greater than 1

C. Parameter 'n' should be a float, not int

D. Missing import for numpy

Solution

1. Step 1: Check parameter types for rvs

   The size parameter expects an integer or tuple of integers, not a string.

2. Step 2: Analyze given parameters

   Here, size='5' is a string, which will cause a TypeError.

3. Final Answer:

   Parameter 'size' should be an integer, not a string -> Option A

4. Quick Check:

   size must be int, not string [OK]

Hint: Size must be int or tuple, never string [OK]

Common Mistakes:

• Passing size as string instead of int
• Thinking p must be >1 (it's a probability 0-1)
• Assuming n must be float

5. You want to simulate the number of successes in 20 trials where each trial has a 30% chance of success. Which scipy.stats function call correctly generates 10 such random samples?

hard

A. binomial.rvs(20, 0.3, 10)

B. binom.rvs(trials=20, prob=0.3, samples=10)

C. binom.rvs(n=20, p=0.3, size=10)

D. binom.sample(n=20, p=0.3, size=10)

Solution

1. Step 1: Identify correct distribution and parameters

   The binomial distribution models number of successes in fixed trials. Parameters are n for trials, p for success probability, and size for sample count.

2. Step 2: Check each option's correctness

   Only binom.rvs(n=20, p=0.3, size=10) uses correct function binom.rvs and correct parameter names.

3. Final Answer:

   binom.rvs(n=20, p=0.3, size=10) -> Option C

4. Quick Check:

   binom.rvs with n, p, size correct = binom.rvs(n=20, p=0.3, size=10) [OK]

Hint: Use binom.rvs with n, p, size for binomial samples [OK]

Common Mistakes:

• Using wrong parameter names like trials or prob
• Using non-existent functions like binomial.rvs
• Confusing sample with rvs method