Random sampling distributions in NumPy - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a random sampling distribution?

A random sampling distribution shows how a statistic (like the mean) varies when we take many samples from the same population. It helps us understand the spread and behavior of that statistic.

beginner

How do you generate random samples using numpy?

You can use numpy's functions like numpy.random.choice() to pick random samples from data, or numpy.random.normal() to generate samples from a normal distribution.

beginner

Why do we use many samples to create a sampling distribution?

Using many samples helps us see the range and pattern of the statistic's values. This shows how much the statistic can change just by chance.

intermediate

What does the Central Limit Theorem say about sampling distributions?

It says that if you take many samples of a large enough size, the distribution of the sample means will look like a normal (bell-shaped) curve, no matter the original data shape.

beginner

How can you visualize a sampling distribution in Python?

You can collect many sample statistics (like means) in a list or array, then use a histogram plot (e.g., with matplotlib) to see the shape and spread of the sampling distribution.

Which numpy function can generate random samples from a normal distribution?

A numpy.random.shuffle()

B numpy.random.choice()

C numpy.random.normal()

D numpy.random.randint()

numpy.random.normal() generates samples from a normal distribution. The others serve different purposes.

What does a sampling distribution represent?

A The distribution of a statistic from many samples

B The distribution of the original data

C The distribution of a single sample

D The distribution of random numbers

A sampling distribution shows how a statistic (like the mean) varies across many samples.

According to the Central Limit Theorem, the sampling distribution of the mean will be approximately normal if:

A Population is normal

B Sample size is large

C Sample size is small

D Population is uniform

The Central Limit Theorem states that with a large sample size, the sampling distribution of the mean is approximately normal regardless of population shape.

Which of these is NOT a reason to use random sampling?

A To estimate population parameters

B To get a representative sample

C To reduce bias

D To guarantee the sample mean equals population mean

Random sampling helps reduce bias and estimate parameters but does not guarantee the sample mean equals the population mean.

What Python library is commonly used to visualize sampling distributions?

A matplotlib

B numpy

C pandas

D scipy

matplotlib is commonly used to create histograms and plots to visualize sampling distributions.

Explain what a random sampling distribution is and why it is important in statistics.

Think about how sample means change when you take many samples.

Describe how you would use numpy to create a sampling distribution of the mean from a dataset.

Imagine picking many small groups from your data and finding their averages.

Practice

1. What does the numpy.random.choice function do in random sampling?

easy

A. It selects random elements from a given array or list.

B. It generates random numbers from a normal distribution.

C. It creates a sequence of numbers with equal spacing.

D. It calculates the mean of a dataset.

Solution

1. Step 1: Understand the purpose of numpy.random.choice

   This function is used to randomly pick elements from an input array or list, optionally with replacement.

2. Step 2: Compare with other options

   Options A, B, and C describe different numpy functions or operations unrelated to random selection from an array.

3. Final Answer:

   It selects random elements from a given array or list. -> Option A

4. Quick Check:

   numpy.random.choice = random selection [OK]

Hint: Remember: choice picks from arrays, normal generates bell curve numbers [OK]

Common Mistakes:

• Confusing choice with normal distribution generator
• Thinking choice creates sequences
• Assuming choice calculates statistics

2. Which of the following is the correct syntax to generate 5 samples from a binomial distribution with 10 trials and success probability 0.3 using numpy?

easy

A. numpy.binomial(10, 0.3, 5)

B. numpy.random.binomial(10, 0.3, size=5)

C. numpy.random.binomial(size=5, 10, 0.3)

D. numpy.random.binomial(5, 10, 0.3)

Solution

1. Step 1: Recall the numpy binomial function signature

   The correct syntax is numpy.random.binomial(n, p, size) where n is number of trials, p is success probability, and size is number of samples.

2. Step 2: Check each option

   numpy.random.binomial(10, 0.3, size=5) matches the correct order and uses the keyword size. numpy.binomial(10, 0.3, 5) misses .random. numpy.random.binomial(size=5, 10, 0.3) mixes keyword size=5 followed by positionals, causing SyntaxError. numpy.random.binomial(5, 10, 0.3) swaps parameters incorrectly.

3. Final Answer:

   numpy.random.binomial(10, 0.3, size=5) -> Option B

4. Quick Check:

   binomial(n, p, size) = numpy.random.binomial(10, 0.3, size=5) [OK]

Hint: Remember order: trials, probability, then size keyword [OK]

Common Mistakes:

• Mixing argument order
• Omitting size keyword
• Using wrong function name

3. What is the output of the following code?

import numpy as np
np.random.seed(0)
samples = np.random.normal(loc=5, scale=2, size=3)
print(np.round(samples, 2))

medium

A. [7.65 5.12 6.34]

B. [8.53 4.32 6.58]

C. [8.82 4.0 6.89]

D. [8.53 5.8 6.96]

Solution

1. Step 1: Understand the code

   The code sets a random seed for reproducibility, then generates 3 samples from a normal distribution with mean 5 and standard deviation 2.

2. Step 2: Run or recall the output

   Using seed 0, the samples generated are [8.53, 5.8, 6.96] after rounding to 2 decimals.

3. Final Answer:

   [8.53 5.8 6.96] -> Option D

4. Quick Check:

   Seed 0 normal samples = [8.53 5.8 6.96] [OK]

Hint: Use seed for fixed output, then round to compare [OK]

Common Mistakes:

• Ignoring seed causes different outputs
• Rounding incorrectly
• Mixing mean and scale parameters

4. Identify the error in this code snippet that tries to sample 4 values from a uniform distribution between 0 and 1:

import numpy as np
samples = np.random.uniform(low=0, high=1, size=4)
print(samples)

medium

A. There is no error; the code runs correctly.

B. The parameters low and high are reversed.

C. The function name should be random.uniform without numpy prefix.

D. The size parameter must be a tuple, not an integer.

Solution

1. Step 1: Review the numpy uniform function syntax

   np.random.uniform(low, high, size) is the correct way to generate samples from a uniform distribution.

2. Step 2: Check the code correctness

   The code uses correct function name, parameter order, and size as integer for 1D array. It will run without error and print 4 random values between 0 and 1.

3. Final Answer:

   There is no error; the code runs correctly. -> Option A

4. Quick Check:

   Uniform sampling syntax correct = There is no error; the code runs correctly. [OK]

Hint: Check function name and parameter order carefully [OK]

Common Mistakes:

• Confusing numpy.random.uniform with Python's random.uniform
• Swapping low and high values
• Using tuple for size unnecessarily

5. You want to simulate rolling a fair six-sided die 1000 times and count how many times the number 4 appears. Which numpy code correctly performs this simulation?

hard

A. rolls = np.random.choice([1,2,3,4,5,6], size=1000)
count_4 = np.sum(rolls == 4)

B. rolls = np.random.binomial(6, 1/6, size=1000)
count_4 = np.count_nonzero(rolls == 4)

C. rolls = np.random.randint(1, 7, size=1000)
count_4 = np.sum(rolls == 4)

D. rolls = np.random.normal(4, 1, size=1000)
count_4 = np.sum(rolls == 4)

Solution

1. Step 1: Understand the problem

   We want to simulate rolling a die 1000 times, each roll equally likely from 1 to 6, then count how many times 4 appears.

2. Step 2: Evaluate each option

   rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4): rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) uses randint from 1 to 6 (exclusive upper bound 7), perfect for die rolls. rolls = np.random.binomial(6, 1/6, size=1000) count_4 = np.count_nonzero(rolls == 4): rolls = np.random.binomial(6, 1/6, size=1000) count_4 = np.count_nonzero(rolls == 4) uses binomial which models number of successes in trials, not die rolls. rolls = np.random.choice([1,2,3,4,5,6], size=1000) count_4 = np.sum(rolls == 4): rolls = np.random.choice([1,2,3,4,5,6], size=1000) count_4 = np.sum(rolls == 4) omits replace=True (defaults to False), raising ValueError since size=1000 > 6. rolls = np.random.normal(4, 1, size=1000) count_4 = np.sum(rolls == 4): rolls = np.random.normal(4, 1, size=1000) count_4 = np.sum(rolls == 4) uses normal distribution, inappropriate for discrete die rolls.

3. Step 3: Choose the best correct code

   rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) is the correct way to simulate die rolls and count 4s.

4. Final Answer:

   rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) -> Option C

5. Quick Check:

   Use randint for discrete uniform die rolls = rolls = np.random.randint(1, 7, size=1000) count_4 = np.sum(rolls == 4) [OK]

Hint: Use randint(1,7) for dice rolls, then count matches [OK]

Common Mistakes:

• Using binomial instead of uniform discrete
• Using normal distribution for discrete outcomes
• Forgetting replace=True in choice