The standard error of the estimate of a proportion $\hat{p}$ is given by the formula:

$SE(\hat{p}) = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

where:

• $\hat{p}$ is the sample proportion (in this case, 0.2),
• $1 - \hat{p}$ is the proportion of individuals who do not like tapioca pudding (0.8),
• $n$ is the sample size (100).

Substituting the values into the formula:

$SE(\hat{p}) = \sqrt{\frac{0.2 \times 0.8}{100}} = \sqrt{\frac{0.16}{100}} = \sqrt{0.0016} = 0.04$

Thus, the standard error of the estimate of the proportion of individuals who like tapioca pudding is 0.04.

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the sample size affect the standard error of the proportion?
2. What would the standard error be if $\hat{p}$ was 0.5?
3. How can we use this standard error to calculate a confidence interval for the proportion?
4. What is the relationship between standard error and margin of error?
5. How would the standard error change if we increased the sample size to 200?

Tip: Increasing the sample size decreases the standard error, improving the precision of your estimate.