Uniform distribution in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of uniform random sample generation

What is the output of this code snippet that generates 3 random numbers from a uniform distribution between 0 and 1?

SciPy

from scipy.stats import uniform
sample = uniform.rvs(size=3, random_state=42)
print([round(x, 2) for x in sample])

A [0.37, 0.95, 0.73, 0.60]

B [0.37, 0.95, 0.73]

C [0.42, 0.58, 0.91]

D [0.73, 0.37, 0.95]

💡 Hint

Use the random_state to get reproducible results and round the output to 2 decimals.

The uniform.rvs function with random_state=42 produces the same 3 values every time. Rounded to 2 decimals, they are [0.37, 0.95, 0.73].

❓ data_output

intermediate

Mean and variance of uniform distribution

What are the mean and variance of a uniform distribution between 2 and 5?

SciPy

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

A 3.0 1.0

B 3.5 1.0

C 2.5 0.75

D 3.5 0.75

💡 Hint

Mean is midpoint; variance is (scale^2)/12.

Mean = (2 + 5)/2 = 3.5; Variance = (3^2)/12 = 9/12 = 0.75.

❓ visualization

advanced

Plotting PDF of uniform distribution

Which option produces the correct plot of the probability density function (PDF) of a uniform distribution from 0 to 4?

SciPy

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import uniform
x = np.linspace(-1, 5, 500)
rv = uniform(loc=0, scale=4)
plt.plot(x, rv.pdf(x))
plt.title('Uniform PDF from 0 to 4')
plt.show()

A A flat line at height 0.25 between 0 and 4, zero elsewhere

B A bell-shaped curve centered at 2

C A line increasing from 0 to 4

D A line decreasing from 0 to 4

💡 Hint

Uniform PDF is constant between loc and loc+scale.

The PDF of uniform(0,4) is 1/4=0.25 between 0 and 4, zero outside. So the plot is a flat line at 0.25 in that range.

🧠 Conceptual

advanced

Effect of scale parameter on uniform distribution

If you increase the scale parameter of a uniform distribution while keeping the location fixed, what happens to the variance?

A Variance stays the same

B Variance decreases linearly with scale

C Variance increases quadratically with scale

D Variance increases linearly with scale

💡 Hint

Variance formula involves square of scale.

Variance of uniform(loc, scale) is (scale^2)/12, so it grows with the square of scale.

🔧 Debug

expert

Identify the error in uniform distribution sampling code

What error does this code raise? from scipy.stats import uniform sample = uniform.rvs(loc=1, scale=-2, size=5) print(sample)

SciPy

from scipy.stats import uniform
sample = uniform.rvs(loc=1, scale=-2, size=5)
print(sample)

A ValueError: scale parameter must be positive

B TypeError: size must be an integer

C No error, prints 5 samples

D RuntimeWarning: invalid value encountered

💡 Hint

Scale parameter cannot be negative for uniform distribution.

The scale parameter must be positive; negative scale raises ValueError.

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