Probability distributions in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Normal Distribution PDF Calculation

What is the output of this MATLAB code snippet that calculates the probability density function (PDF) of a normal distribution at x = 0?

MATLAB

mu = 0;
sigma = 1;
x = 0;
pdf_val = normpdf(x, mu, sigma);
disp(pdf_val);

A 0.3989

B 0

C 1

D NaN

💡 Hint

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

The normal distribution PDF at the mean (0) with standard deviation 1 is 1/sqrt(2*pi) ≈ 0.3989.

❓ Predict Output

intermediate

Result of Binomial Distribution CDF Calculation

What is the output of this MATLAB code that calculates the cumulative distribution function (CDF) of a binomial distribution with n=5 trials and probability p=0.5 at x=3?

MATLAB

n = 5;
p = 0.5;
x = 3;
cdf_val = binocdf(x, n, p);
disp(cdf_val);

A 0.8125

B 0.625

C 0.5

D 1

💡 Hint

Sum probabilities of 0 to 3 successes in binomial distribution.

binocdf(3,5,0.5) sums probabilities of 0,1,2,3 successes, which equals 0.8125.

🔧 Debug

advanced

Identify the Error in Poisson PMF Calculation

What error does this MATLAB code produce when calculating the Poisson probability mass function (PMF) for k=3 with lambda=2?

MATLAB

lambda = 2;
k = 3;
pmf_val = poisspdf(k, lambda
); disp(pmf_val);

A RuntimeError: Undefined function 'poisspdf'

B SyntaxError: Missing closing parenthesis

C ValueError: k must be integer

D No error, outputs 0.1804

💡 Hint

Check the parentheses in the function call.

The code is missing a closing parenthesis in the poisspdf function call, causing a syntax error.

❓ Predict Output

advanced

Output of Exponential Distribution Random Sample Mean

What is the approximate output of this MATLAB code that generates 1000 samples from an exponential distribution with mean 1 and calculates their mean?

MATLAB

lambda = 1;
samples = exprnd(1/lambda, 1000, 1);
mean_val = mean(samples);
disp(round(mean_val,2));

A 2.00

B 0.50

C 1.00

D NaN

💡 Hint

The mean of exponential distribution is 1/lambda.

The sample mean should be close to the theoretical mean 1 because of the law of large numbers.

🧠 Conceptual

expert

Which Option Correctly Describes the Relationship Between PDF and CDF?

Which of the following statements correctly describes the relationship between the probability density function (PDF) and the cumulative distribution function (CDF) for a continuous random variable?

A The PDF is the derivative of the CDF with respect to x only when the variable is discrete.

B The PDF is the integral of the CDF from negative infinity to x.

C The PDF and CDF are always equal for all x.

D The CDF is the integral of the PDF from negative infinity to x.

💡 Hint

Think about how area under the PDF curve relates to the CDF.

For continuous variables, the CDF at x is the area under the PDF curve from -∞ to x, so CDF is the integral of PDF.

Practice

1. Which MATLAB function is used to create a probability distribution object?

easy

A. rand

B. makedist

C. histogram

D. plot

Solution

1. Step 1: Understand the purpose of each function

   makedist creates a probability distribution object, while rand generates random numbers, histogram plots data distribution, and plot creates graphs.

2. Step 2: Identify the correct function for creating distributions

   Only makedist is designed to create probability distribution objects in MATLAB.

3. Final Answer:

   makedist -> Option B

4. Quick Check:

   Function to create distribution = makedist [OK]

Hint: Use makedist to create distributions in MATLAB [OK]

Common Mistakes:

• Confusing makedist with rand
• Using histogram to create distributions
• Trying to use plot for distributions

2. Which of the following is the correct syntax to create a normal distribution with mean 0 and standard deviation 1 in MATLAB?

easy

A. pd = makedist('normal', 0, 1);

B. pd = makedist('Normal', 'mean', 0, 'std', 1);

C. pd = makedist('Normal', 0, 1);

D. pd = makedist('Normal', 'mu', 0, 'sigma', 1);

Solution

1. Step 1: Recall the correct parameter names for normal distribution

   In MATLAB, the normal distribution parameters are specified as 'mu' for mean and 'sigma' for standard deviation.

2. Step 2: Check the syntax for makedist

   The correct syntax uses name-value pairs: makedist('Normal', 'mu', 0, 'sigma', 1). Other options use incorrect parameter names or formats.

3. Final Answer:

   pd = makedist('Normal', 'mu', 0, 'sigma', 1); -> Option D

4. Quick Check:

   Normal dist params = 'mu' and 'sigma' [OK]

Hint: Use 'mu' and 'sigma' as parameter names for normal dist [OK]

Common Mistakes:

• Using lowercase 'normal' instead of 'Normal'
• Using 'mean' and 'std' instead of 'mu' and 'sigma'
• Passing parameters without name-value pairs

3. What is the output of the following MATLAB code?

pd = makedist('Exponential', 'mu', 2);
p = pdf(pd, 1);
disp(round(p, 3));

medium

A. 0.303

B. 0.368

C. 0.500

D. 0.607

Solution

1. Step 1: Understand the exponential distribution PDF formula

   The PDF for exponential with mean mu=2 is f(x) = (1/2)*exp(-x/2). For x=1, f(1) = 0.5 * exp(-0.5) ≈ 0.5 * 0.6065 = 0.3033.

2. Step 2: Compute and round the value

   p ≈ 0.3033, round(p, 3) = 0.303.

3. Final Answer:

   0.303 -> Option A

4. Quick Check:

   (1/mu)*exp(-x/mu) at x=1, mu=2 ≈ 0.303 [OK]

Hint: Recall pdf = (1/mu)*exp(-x/mu) for exponential [OK]

Common Mistakes:

• Ignoring the (1/mu) factor
• Confusing PDF with CDF
• Rounding errors causing mismatch

4. Identify the error in this MATLAB code snippet:

pd = makedist('Binomial', 'n', 10, 'p', 0.5);
prob = cdf(pd, 11);

medium

A. The probability p must be an integer

B. makedist does not support 'Binomial' distribution

C. The cdf input value 11 exceeds the number of trials n

D. The variable name 'prob' is reserved

Solution

1. Step 1: Understand binomial distribution parameters

   Binomial distribution with n=10 means possible outcomes are from 0 to 10.

2. Step 2: Check the cdf input value

   Calling cdf(pd, 11) asks for probability of outcomes ≤ 11, but 11 is outside the valid range (max 10).

3. Step 3: Identify the error

   Input to cdf must be within 0 to n. Using 11 is invalid and causes an error or unexpected result.

4. Final Answer:

   The cdf input value 11 exceeds the number of trials n -> Option C

5. Quick Check:

   cdf input must be ≤ n for binomial [OK]

Hint: cdf input must be within distribution support range [OK]

Common Mistakes:

• Using cdf input outside valid range
• Thinking p must be integer
• Assuming makedist lacks binomial support

5. You want to calculate the probability that a normally distributed variable with mean 5 and standard deviation 2 is less than 7 in MATLAB using a probability distribution object. Which code correctly computes this?

hard

A. pd = makedist('Normal', 'mu', 5, 'sigma', 2); p = cdf(pd, 7);

B. p = normcdf(7, 5, 2);

C. pd = makedist('Normal', 5, 2); p = pdf(pd, 7);

D. p = cdf('Normal', 7, 5, 2);

Solution

1. Step 1: Understand how to create a normal distribution object

   Use makedist('Normal', 'mu', 5, 'sigma', 2) to create the distribution with mean 5 and std 2.

2. Step 2: Use cdf to find probability less than 7

   Calling cdf(pd, 7) returns the cumulative probability up to 7.

3. Step 3: Check other options

   p = normcdf(7, 5, 2); uses normcdf which is valid but not using makedist. pd = makedist('Normal', 5, 2); p = pdf(pd, 7); incorrectly uses pdf instead of cdf. p = cdf('Normal', 7, 5, 2); has wrong syntax for cdf.

4. Final Answer:

   pd = makedist('Normal', 'mu', 5, 'sigma', 2); p = cdf(pd, 7); -> Option A

5. Quick Check:

   Use makedist + cdf for cumulative probability [OK]

Hint: Use cdf with makedist object for cumulative probability [OK]

Common Mistakes:

• Using pdf instead of cdf
• Wrong syntax for cdf
• Ignoring distribution parameters