Uniform distribution in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the uniform distribution from scipy.stats.

SciPy

from scipy.stats import [1]

Drag options to blanks, or click blank then click option'

Anorm

Bbinom

Cuniform

Dpoisson

Attempts:

3 left

💡 Hint

Common Mistakes

Importing a different distribution like norm or binom instead of uniform.

The uniform distribution is imported from scipy.stats using from scipy.stats import uniform.

2fill in blank

medium

Complete the code to create a uniform distribution object with lower bound 0 and upper bound 10.

SciPy

dist = uniform(loc=0, scale=[1])

Drag options to blanks, or click blank then click option'

A10

B5

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the upper bound directly as scale without subtracting lower bound.

Setting scale to 0 or 1 incorrectly.

The scale parameter is the width of the uniform distribution interval. For bounds 0 to 10, scale is 10.

3fill in blank

hard

Fix the error in the code to calculate the probability density function (PDF) at x=5 for the uniform distribution.

SciPy

pdf_value = dist.[1](5)

Drag options to blanks, or click blank then click option'

Acdf

Bppf

Cpmf

Dpdf

Attempts:

3 left

💡 Hint

Common Mistakes

Using pmf which is for discrete distributions.

Using cdf instead of pdf.

The PDF (probability density function) is calculated using the pdf method for continuous distributions like uniform.

4fill in blank

hard

Fill both blanks to create a dictionary of probabilities for values 2 and 8 using the PDF method.

SciPy

probabilities = {2: dist.[1](2), 8: dist.[2](8)}

Drag options to blanks, or click blank then click option'

Apdf

Bcdf

Cpmf

Dppf

Attempts:

3 left

💡 Hint

Common Mistakes

Using different methods for each value.

Using PMF or CDF instead of PDF.

Use the pdf method to get the probability density at specific points for a uniform distribution.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each number in the list to its PDF value if the number is greater than 3.

SciPy

result = {num: dist.[1](num) for num in numbers if num [2] 3 and num [3] 10}

Drag options to blanks, or click blank then click option'

Apdf

B>

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators.

Using PMF or CDF instead of PDF.

The dictionary comprehension uses pdf to get probabilities. The condition filters numbers greater than 3 and less than 10.

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