Uniform distribution in SciPy - Step-by-Step Execution

Concept Flow - Uniform distribution

Set parameters a and b

↓

Generate random values

↓

Calculate PDF or CDF

↓

Analyze or visualize results

↓

End

The flow starts by setting the range parameters a and b, then generates random values from the uniform distribution, calculates probabilities or cumulative values, and finally analyzes or visualizes the results.

Execution Sample

SciPy

from scipy.stats import uniform

# Define range
rv = uniform(loc=0, scale=5)

# Generate 5 random samples
samples = rv.rvs(size=5)

# Calculate PDF at 2.5
pdf_val = rv.pdf(2.5)

# Calculate CDF at 2.5
cdf_val = rv.cdf(2.5)

This code creates a uniform distribution from 0 to 5, generates 5 random samples, and calculates the probability density and cumulative distribution at 2.5.

Execution Table

Step	Action	Value/Expression	Result/Output
1	Create uniform distribution with loc=0, scale=5	uniform(loc=0, scale=5)	Distribution object rv created
2	Generate 5 random samples	rv.rvs(size=5)	[3.1, 0.7, 4.8, 2.3, 1.5] (example)
3	Calculate PDF at 2.5	rv.pdf(2.5)	0.2
4	Calculate CDF at 2.5	rv.cdf(2.5)	0.5
5	Calculate PDF at 6 (outside range)	rv.pdf(6)	0.0
6	Calculate CDF at -1 (below range)	rv.cdf(-1)	0.0
7	Calculate CDF at 5 (upper bound)	rv.cdf(5)	1.0
8	End	-	Execution complete

💡 All steps executed to demonstrate sampling and probability calculations within and outside the uniform distribution range.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
rv	None	uniform(loc=0, scale=5)	uniform(loc=0, scale=5)	uniform(loc=0, scale=5)	uniform(loc=0, scale=5)	uniform(loc=0, scale=5)	uniform(loc=0, scale=5)	uniform(loc=0, scale=5)
samples	None	[3.1, 0.7, 4.8, 2.3, 1.5]	[3.1, 0.7, 4.8, 2.3, 1.5]	[3.1, 0.7, 4.8, 2.3, 1.5]	[3.1, 0.7, 4.8, 2.3, 1.5]	[3.1, 0.7, 4.8, 2.3, 1.5]	[3.1, 0.7, 4.8, 2.3, 1.5]	[3.1, 0.7, 4.8, 2.3, 1.5]
pdf_val	None	None	0.2	0.2	0.2	0.2	0.2	0.2
cdf_val_2_5	None	None	None	0.5	0.5	0.5	0.5	0.5
pdf_val_6	None	None	None	None	0.0	0.0	0.0	0.0
cdf_val_neg1	None	None	None	None	None	0.0	0.0	0.0
cdf_val_5	None	None	None	None	None	None	1.0	1.0

Key Moments - 3 Insights

Why is the PDF value constant inside the range but zero outside?

The uniform distribution has equal probability density between loc and loc+scale, so PDF is constant there (0.2 here). Outside this range, the PDF is zero because values cannot occur there. See execution_table rows 3 and 5.

What does the CDF value represent at a point inside the range?

CDF gives the probability that a value is less than or equal to that point. For 2.5, it is 0.5, meaning 50% chance values are ≤ 2.5. See execution_table row 4.

Why does the CDF at the upper bound equal 1?

CDF at the upper bound (loc+scale) is always 1 because all values in the distribution are less than or equal to that point. See execution_table row 7.

Visual Quiz - 3 Questions

Look at the execution_table row 3. What is the PDF value at 2.5?

A 0.2

B 0.5

C 0.0

D 1.0

💡 Hint

Check the 'Result/Output' column in row 3 for the PDF calculation.

At which step does the CDF value become 1.0?

A Step 4

B Step 7

C Step 5

D Step 6

💡 Hint

Look at the 'CDF at 5' calculation in the execution_table.

If the scale parameter changed from 5 to 10, how would the PDF at 2.5 change?

A It would increase

B It would stay the same

C It would decrease

D It would become zero

💡 Hint

Recall PDF for uniform is 1/scale; increasing scale lowers PDF value.

Concept Snapshot

Uniform distribution syntax:
rv = uniform(loc=a, scale=b-a)
Generates values equally likely between a and b.
PDF is constant: 1/(b - a) inside [a,b], zero outside.
CDF increases linearly from 0 to 1 over [a,b].
Use rv.rvs(size), rv.pdf(x), rv.cdf(x) for sampling and probabilities.

Full Transcript

This visual execution traces the uniform distribution from scipy.stats. We start by creating a uniform distribution object with loc=0 and scale=5, representing values between 0 and 5. Then, we generate 5 random samples from this distribution. Next, we calculate the probability density function (PDF) at 2.5, which is constant inside the range and equals 0.2. We also calculate the cumulative distribution function (CDF) at 2.5, which is 0.5, meaning there's a 50% chance a value is less than or equal to 2.5. We check PDF and CDF values outside the range to see they are zero or one appropriately. The variable tracker shows how each variable changes after each step. Key moments clarify why PDF is constant inside the range and zero outside, what CDF means inside the range, and why CDF at the upper bound is 1. The quiz questions test understanding of these values and effects of changing parameters. The snapshot summarizes how to use the uniform distribution in scipy with key properties.

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