Uniform distribution in SciPy - Mini Project: Build & Apply

Exploring Uniform Distribution with SciPy

📖 Scenario: Imagine you work for a company that wants to understand how to generate random numbers that are equally likely to fall anywhere between two values. This is useful in simulations and games where fairness is important.

🎯 Goal: You will create a simple program to generate random numbers from a uniform distribution using SciPy. You will set the range, generate numbers, and then display the results.

📋 What You'll Learn

Create a uniform distribution with specific lower and upper bounds

Generate random numbers from this distribution

Calculate the mean of the generated numbers

Print the generated numbers and their mean

💡 Why This Matters

🌍 Real World

Uniform distributions are used in simulations, games, and modeling situations where every outcome in a range is equally likely.

💼 Career

Understanding uniform distributions helps in data science roles involving simulations, random sampling, and probabilistic modeling.

1

Create the uniform distribution range

Create two variables called lower_bound and upper_bound with values 10 and 20 respectively.

SciPy

# Your code here

Hint

Use simple assignment to create lower_bound and upper_bound.

2

Set up the uniform distribution object

Import uniform from scipy.stats and create a variable called dist that represents a uniform distribution from lower_bound to upper_bound. Use uniform(loc=lower_bound, scale=upper_bound - lower_bound).

SciPy

lower_bound = 10
upper_bound = 20
# Your code here

Hint

Use uniform(loc=lower_bound, scale=upper_bound - lower_bound) to create the distribution.

3

Generate random numbers from the distribution

Use the rvs method of dist to generate 5 random numbers and store them in a list called random_numbers.

SciPy

from scipy.stats import uniform

lower_bound = 10
upper_bound = 20
dist = uniform(loc=lower_bound, scale=upper_bound - lower_bound)
# Your code here

Hint

Use dist.rvs(size=5) to get 5 random values, then convert to list with tolist().

4

Print the random numbers and their mean

Print the list random_numbers and then print the mean of these numbers using sum(random_numbers) / len(random_numbers).

SciPy

from scipy.stats import uniform

lower_bound = 10
upper_bound = 20
dist = uniform(loc=lower_bound, scale=upper_bound - lower_bound)
random_numbers = dist.rvs(size=5).tolist()
# Your code here

Hint

Use print(random_numbers) and then print(sum(random_numbers) / len(random_numbers)).

Practice

1. What does the loc parameter represent in the scipy.stats.uniform distribution?

easy

A. The starting point (minimum value) of the uniform distribution range

B. The width (length) of the uniform distribution range

C. The mean of the uniform distribution

D. The variance of the uniform distribution

Solution

1. Step 1: Understand the parameters of uniform distribution

   The uniform distribution in scipy uses loc as the starting point of the range and scale as the length of the range.

2. Step 2: Identify the role of loc

   The loc parameter sets the minimum value where the uniform distribution starts.

3. Final Answer:

   The starting point (minimum value) of the uniform distribution range -> Option A

4. Quick Check:

   loc = start value [OK]

Hint: Remember: loc = start, scale = length [OK]

Common Mistakes:

• Confusing loc with scale
• Thinking loc is the mean
• Mixing loc with variance

2. Which of the following is the correct way to create a uniform distribution from 5 to 15 using scipy?

easy

A. uniform(loc=10, scale=5)

B. uniform(loc=5, scale=15)

C. uniform(loc=5, scale=10)

D. uniform(loc=0, scale=10)

Solution

1. Step 1: Calculate scale from range

   The range is from 5 to 15, so the length is 15 - 5 = 10. This is the scale.

2. Step 2: Set loc and scale correctly

   Set loc=5 (start) and scale=10 (length) to cover 5 to 15.

3. Final Answer:

   uniform(loc=5, scale=10) -> Option C

4. Quick Check:

   loc=5, scale=15-5=10 [OK]

Hint: Scale = end - start, loc = start [OK]

Common Mistakes:

• Swapping loc and scale values
• Using scale as the end value
• Setting loc to zero incorrectly

3. What is the output of this code?

from scipy.stats import uniform
rv = uniform(loc=2, scale=3)
print(round(rv.pdf(3), 2))

medium

A. 0.33

B. 0.20

C. 0.50

D. 0.00

Solution

1. Step 1: Understand PDF of uniform distribution

   PDF is constant inside the range and equals 1/scale. Here, scale=3, so PDF = 1/3 ≈ 0.3333.

2. Step 2: Check if x=3 is inside the range

   Range is from loc=2 to 2+3=5. Since 3 is inside, PDF at 3 is 1/3 ≈ 0.33.

3. Final Answer:

   0.33 -> Option A

4. Quick Check:

   PDF = 1/scale = 1/3 = 0.33 [OK]

Hint: PDF inside range = 1/scale [OK]

Common Mistakes:

• Calculating PDF as zero inside range
• Using loc instead of scale for PDF
• Rounding errors without checking range

4. Identify the error in this code snippet:

from scipy.stats import uniform
rv = uniform(loc=1, scale=4)
prob = rv.cdf(6)
print(prob)

medium

A. The scale should be 5 to cover up to 6

B. No error; code runs correctly

C. The loc parameter must be zero

D. The cdf method does not exist for uniform

Solution

1. Step 1: Check parameters and range

   loc=1 and scale=4 means range is from 1 to 5 (1+4=5).

2. Step 2: Evaluate cdf at 6

   6 is outside the range (greater than 5), so cdf(6) should return 1 (full probability).

3. Step 3: Confirm method existence and output

   The cdf method exists for uniform and returns valid output even outside range.

4. Final Answer:

   No error; code runs correctly -> Option B

5. Quick Check:

   cdf exists and returns 1 for x > loc+scale [OK]

Hint: CDF returns 1 for x beyond range max [OK]

Common Mistakes:

• Thinking scale must cover queried x
• Assuming cdf method is missing
• Believing loc must be zero

5. You want to model a uniform distribution for daily temperatures between 10°C and 20°C. Which code correctly calculates the probability that temperature is less than 15°C?

hard

A. rv = uniform(loc=10, scale=5)
prob = rv.cdf(15)

B. rv = uniform(loc=10, scale=20)
prob = rv.cdf(15)

C. rv = uniform(loc=0, scale=10)
prob = rv.cdf(15)

D. rv = uniform(loc=10, scale=10)
prob = rv.cdf(15)

Solution

1. Step 1: Set correct loc and scale for temperature range

   Range is 10 to 20, so loc=10 and scale=20-10=10.

2. Step 2: Calculate CDF at 15

   CDF at 15 is (15 - loc)/scale = (15-10)/10 = 0.5, which is the probability temperature is less than 15.

3. Final Answer:

   rv = uniform(loc=10, scale=10) prob = rv.cdf(15) -> Option D

4. Quick Check:

   prob = (15-10)/10 = 0.5 [OK]

Hint: CDF = (x - loc) / scale inside range [OK]

Common Mistakes:

• Using wrong scale length
• Setting loc to zero incorrectly
• Calculating CDF outside range without adjustment