To solve this problem, we can use the properties of a normal distribution since the data is assumed to be approximately bell-shaped (normally distributed).
Given:
- Mean (μ) = 25 days
- Standard deviation (σ) = 4 days
We want to find the percentage of bills for which payment was made in greater than 33 days.
- First, we need to compute the z-score for 33 days. The z-score is calculated using the formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
where:
- \(X\) is the value we are interested in (33 days),
- \(\mu\) is the mean (25 days),
- \(\sigma\) is the standard deviation (4 days).
Substituting in the values:
\[ z = \frac{(33 - 25)}{4} = \frac{8}{4} = 2 \]
- Now, we can use the z-score to find the corresponding percentage in the standard normal distribution table (or by using a calculator or software that provides cumulative probabilities).
The z-score of 2 corresponds to the cumulative probability of about 0.9772 (or 97.72%). This means that approximately 97.72% of the payments are made in 33 days or less.
- To find the percentage of bills for which payment was made in greater than 33 days, we subtract this value from 100%:
\[ P(X > 33) = 1 - P(X < 33) \] \[ P(X > 33) = 1 - 0.9772 = 0.0228 \]
- Finally, convert this probability to a percentage:
\[ P(X > 33) \times 100 = 0.0228 \times 100 = 2.28% \]
So, approximately 2.28% of the bills are paid in greater than 33 days.