Probability distributions in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a normal distribution with mean 0 and standard deviation 1.

MATLAB

pd = makedist('Normal', 'mu', [1], 'sigma', 1);

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

A1

B0

C-1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using standard deviation value instead of mean.

Confusing mean with sigma.

The mean (mu) of the normal distribution is 0, so we fill in 0.

2fill in blank

medium

Complete the code to calculate the probability density function (pdf) at x = 2 for the distribution pd.

MATLAB

y = pdf(pd, [1]);

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

A2

B1

C0

D-2

Attempts:

3 left

💡 Hint

Common Mistakes

Using the mean instead of the point 2.

Passing a vector instead of a scalar.

We want the pdf value at x = 2, so we fill in 2.

3fill in blank

hard

Fix the error in the code to generate 5 random samples from the distribution pd.

MATLAB

samples = random(pd, [1], 1);

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

A1

B10

C0

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 5, which generates only one sample.

Using 0 or negative numbers.

To generate 5 samples, the first dimension should be 5.

4fill in blank

hard

Fill both blanks to create a binomial distribution with 10 trials and success probability 0.3.

MATLAB

pd = makedist('Binomial', 'N', [1], 'p', [2]);

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

A10

B0.3

C5

D0.5

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping N and p values.

Using invalid probability values greater than 1.

Number of trials N is 10 and probability p is 0.3.

5fill in blank

hard

Fill all three blanks to create a Poisson distribution with mean 4, calculate the pdf at x=3, and generate 4 random samples.

MATLAB

pd = makedist('Poisson', 'lambda', [1]);
y = pdf(pd, [2]);
samples = random(pd, [3], 1);

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

A3

B4

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up lambda and sample count.

Using wrong values for pdf evaluation.

Lambda (mean) is 4, pdf evaluated at 3, and 4 samples generated.

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