Normal distribution in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Normal Distribution PDF Calculation

What is the output of this code that calculates the probability density function (PDF) of a normal distribution at x=0 with mean=0 and standard deviation=1?

SciPy

from scipy.stats import norm
result = norm.pdf(0, loc=0, scale=1)
print(round(result, 4))

A 0.5000

B 0.3989

C 1.0000

D 0.0000

💡 Hint

Recall the formula for the normal distribution PDF at the mean.

The PDF of a standard normal distribution at its mean (0) is 1 divided by the square root of 2π, approximately 0.3989.

❓ data_output

intermediate

Number of Samples within One Standard Deviation

Using scipy, generate 1000 samples from a normal distribution with mean=5 and std=2. How many samples fall between 3 and 7 (within one standard deviation)?

SciPy

import numpy as np
from scipy.stats import norm
samples = norm.rvs(loc=5, scale=2, size=1000, random_state=42)
count = np.sum((samples >= 3) & (samples <= 7))
print(count)

A 950

B 500

C 682

D 840

💡 Hint

Recall the empirical rule for normal distributions.

About 68% of data in a normal distribution lies within one standard deviation from the mean. 68% of 1000 is about 680.

❓ visualization

advanced

Plotting Normal Distribution PDF with Different Parameters

Which option produces a plot showing two normal distribution curves: one with mean=0, std=1 and another with mean=2, std=0.5?

SciPy

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm
x = np.linspace(-3, 5, 500)
plt.plot(x, norm.pdf(x, 0, 1), label='mean=0, std=1')
plt.plot(x, norm.pdf(x, 2, 0.5), label='mean=2, std=0.5')
plt.legend()
plt.show()

A Plots two curves: a wider bell centered at 0 and a narrower bell centered at 2

B Plots two identical curves both centered at 0

C Plots one curve only centered at 1

D Raises a TypeError due to wrong function usage

💡 Hint

Check the parameters passed to norm.pdf for each curve.

The code plots two normal PDFs with different means and std deviations, resulting in two distinct bell curves.

🧠 Conceptual

advanced

Understanding the Effect of Standard Deviation on Normal Distribution Shape

Which statement correctly describes how increasing the standard deviation affects the shape of a normal distribution?

A The curve becomes wider and shorter, spreading data over a larger range

B The curve becomes narrower and taller, concentrating data near the mean

C The curve shifts to the right without changing shape

D The curve becomes asymmetric, skewing to one side

💡 Hint

Think about how spread relates to standard deviation.

Increasing standard deviation spreads data out, making the bell curve wider and lower in height.

🔧 Debug

expert

Identify the Error in Normal Distribution Sampling Code

What error does this code raise when trying to generate samples from a normal distribution? import numpy as np from scipy.stats import norm samples = norm.rvs(loc=0, scale=-1, size=100) print(samples)

SciPy

import numpy as np
from scipy.stats import norm
samples = norm.rvs(loc=0, scale=-1, size=100)
print(samples)

A RuntimeWarning: invalid value encountered in sqrt

B No error, prints 100 samples

C TypeError: unsupported operand type(s) for -: 'int' and 'str'

D ValueError: scale parameter must be positive

💡 Hint

Check the scale parameter value passed to norm.rvs.

The scale (standard deviation) must be positive. Negative scale causes ValueError.

Practice

1. What does the normal distribution model in data science?

easy

A. Data with no variation

B. Data with only positive values

C. Data sorted in ascending order

D. Data spread around an average value

Solution

1. Step 1: Understand the normal distribution concept

   The normal distribution describes how data points are spread around a central average (mean).

2. Step 2: Identify the correct description

   Among the options, only data spread around an average matches the normal distribution's purpose.

3. Final Answer:

   Data spread around an average value -> Option D

4. Quick Check:

   Normal distribution = data around average [OK]

Hint: Normal distribution centers data around mean [OK]

Common Mistakes:

• Thinking normal distribution only has positive values
• Confusing normal distribution with sorted data
• Assuming no variation in normal distribution

2. Which of the following is the correct way to import the normal distribution functions from scipy.stats?

easy

A. import scipy.norm

B. from scipy.stats import norm

C. from scipy import normal

D. import stats.norm

Solution

1. Step 1: Recall scipy.stats import syntax

   The normal distribution functions are in scipy.stats under the name 'norm'.

2. Step 2: Match correct import statement

   Only 'from scipy.stats import norm' correctly imports the normal distribution functions.

3. Final Answer:

   from scipy.stats import norm -> Option B

4. Quick Check:

   Correct import = from scipy.stats import norm [OK]

Hint: Use 'from scipy.stats import norm' to access normal distribution [OK]

Common Mistakes:

• Using 'import scipy.norm' which does not exist
• Trying 'from scipy import normal' which is incorrect
• Importing 'stats.norm' without scipy prefix

3. What is the output of this code?

from scipy.stats import norm
result = norm.cdf(0, loc=0, scale=1)
print(round(result, 2))

medium

A. 1.00

B. 0.00

C. 0.50

D. 0.84

Solution

1. Step 1: Understand norm.cdf at mean

   norm.cdf(x, loc, scale) gives the probability data is less than x. At x=0 with mean=0, scale=1, it is the 50th percentile.

2. Step 2: Calculate and round the result

   The cumulative probability at mean is 0.5, rounding to 2 decimals gives 0.50.

3. Final Answer:

   0.50 -> Option C

4. Quick Check:

   norm.cdf(mean) = 0.5 [OK]

Hint: CDF at mean of normal = 0.5 [OK]

Common Mistakes:

• Confusing PDF with CDF values
• Expecting 0 or 1 instead of 0.5 at mean
• Rounding errors ignoring decimals

4. Find the error in this code snippet:

from scipy.stats import norm
prob = norm.pdf(1, mean=0, scale=1)
print(prob)

medium

A. Parameter 'mean' should be 'loc'

B. Missing import statement

C. norm.pdf does not exist

D. scale must be an integer

Solution

1. Step 1: Check parameter names for norm.pdf

   norm.pdf uses 'loc' for mean, not 'mean'. Using 'mean' causes an error.

2. Step 2: Identify correct parameter

   Replace 'mean=0' with 'loc=0' to fix the code.

3. Final Answer:

   Parameter 'mean' should be 'loc' -> Option A

4. Quick Check:

   Use loc for mean in norm functions [OK]

Hint: Use 'loc' not 'mean' for normal distribution mean [OK]

Common Mistakes:

• Using 'mean' instead of 'loc' parameter
• Assuming norm.pdf is missing
• Thinking scale must be integer

5. You want to find the probability that a value from a normal distribution with mean 10 and standard deviation 2 is less than 13. Which code correctly calculates this using scipy?

hard

A. norm.cdf(13, loc=10, scale=2)

B. norm.pdf(13, loc=10, scale=2)

C. norm.cdf(10, loc=13, scale=2)

D. norm.sf(13, loc=10, scale=2)

Solution

1. Step 1: Identify the function for cumulative probability

   To find probability less than a value, use norm.cdf with the value, mean (loc), and std dev (scale).

2. Step 2: Match parameters to the problem

   Value is 13, mean is 10, std dev is 2, so norm.cdf(13, loc=10, scale=2) is correct.

3. Final Answer:

   norm.cdf(13, loc=10, scale=2) -> Option A

4. Quick Check:

   Use norm.cdf(value, loc=mean, scale=std) [OK]

Hint: Use norm.cdf for probability less than a value [OK]

Common Mistakes:

• Using norm.pdf instead of norm.cdf
• Swapping loc and value parameters
• Using norm.sf which gives probability greater than value