Question
In a large sample of customer accounts, a utility company determined that the average number of days between when a bill was sent out and when the payment was made is 27 with a standard deviation of 7 days. Assume the data to be approximately bell-shaped.
(a) Between what two values will approximately 68% of the numbers of days be?
Approximately 68% of the customer accounts have payment made between
(a) Between what two values will approximately 68% of the numbers of days be?
Approximately 68% of the customer accounts have payment made between
Answers
GPT-4o mini
In a normal distribution (bell-shaped), approximately 68% of the data falls within one standard deviation of the mean.
Given:
- Mean (average) number of days = 27 days
- Standard deviation = 7 days
To find the range for approximately 68% of the days:
1. Lower value = Mean - Standard deviation = 27 - 7 = 20 days
2. Upper value = Mean + Standard deviation = 27 + 7 = 34 days
Thus, approximately 68% of the customer accounts have payments made between **20 days and 34 days**.
Given:
- Mean (average) number of days = 27 days
- Standard deviation = 7 days
To find the range for approximately 68% of the days:
1. Lower value = Mean - Standard deviation = 27 - 7 = 20 days
2. Upper value = Mean + Standard deviation = 27 + 7 = 34 days
Thus, approximately 68% of the customer accounts have payments made between **20 days and 34 days**.
(b) Estimate the percentage of customer accounts for which the number of days is between 13 and 41
GPT-4o mini
To estimate the percentage of customer accounts for which the number of days between when a bill was sent out and when the payment was made is between 13 and 41 days, we can again use the properties of the normal distribution.
Given:
- Mean = 27 days
- Standard deviation = 7 days
First, we need to determine how many standard deviations away from the mean the values 13 and 41 are:
1. **For the lower value (13 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(13 - 27)}{7} = \frac{-14}{7} = -2
\]
2. **For the upper value (41 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(41 - 27)}{7} = \frac{14}{7} = 2
\]
Now, we look up these \( z \)-scores in the standard normal distribution.
- A \( z \)-score of -2 corresponds to approximately 2.5% in the lower tail.
- A \( z \)-score of 2 corresponds to approximately 97.5% in the upper tail.
To find the percentage of customer accounts within this range:
\[
\text{Percentage} = \text{Upper} - \text{Lower} = 97.5\% - 2.5\% = 95\%
\]
Thus, approximately **95%** of customer accounts have payments made between **13 days and 41 days**.
Given:
- Mean = 27 days
- Standard deviation = 7 days
First, we need to determine how many standard deviations away from the mean the values 13 and 41 are:
1. **For the lower value (13 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(13 - 27)}{7} = \frac{-14}{7} = -2
\]
2. **For the upper value (41 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(41 - 27)}{7} = \frac{14}{7} = 2
\]
Now, we look up these \( z \)-scores in the standard normal distribution.
- A \( z \)-score of -2 corresponds to approximately 2.5% in the lower tail.
- A \( z \)-score of 2 corresponds to approximately 97.5% in the upper tail.
To find the percentage of customer accounts within this range:
\[
\text{Percentage} = \text{Upper} - \text{Lower} = 97.5\% - 2.5\% = 95\%
\]
Thus, approximately **95%** of customer accounts have payments made between **13 days and 41 days**.
(c) Estimate the percentage of customer accounts for which the number of days is between 6 and 48
GPT-4o mini
To estimate the percentage of customer accounts for which the number of days is between 6 and 48, we will again use the properties of the normal distribution.
Given:
- Mean (\( \mu \)) = 27 days
- Standard deviation (\( \sigma \)) = 7 days
We first calculate the z-scores for the values 6 and 48.
1. **For the lower value (6 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(6 - 27)}{7} = \frac{-21}{7} = -3
\]
2. **For the upper value (48 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(48 - 27)}{7} = \frac{21}{7} = 3
\]
Now, we look up these \( z \)-scores in the standard normal distribution:
- A \( z \)-score of -3 corresponds to approximately 0.13% (or about 0.0013 in probability) in the lower tail.
- A \( z \)-score of 3 corresponds to approximately 99.87% (or about 0.9987 in probability) in the upper tail.
To find the percentage of customer accounts within this range, we calculate:
\[
\text{Percentage} = \text{Upper} - \text{Lower} = 99.87\% - 0.13\% \approx 99.74\%
\]
Thus, approximately **99.74%** of customer accounts have payments made between **6 days and 48 days**.
Given:
- Mean (\( \mu \)) = 27 days
- Standard deviation (\( \sigma \)) = 7 days
We first calculate the z-scores for the values 6 and 48.
1. **For the lower value (6 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(6 - 27)}{7} = \frac{-21}{7} = -3
\]
2. **For the upper value (48 days)**:
\[
z = \frac{(X - \mu)}{\sigma} = \frac{(48 - 27)}{7} = \frac{21}{7} = 3
\]
Now, we look up these \( z \)-scores in the standard normal distribution:
- A \( z \)-score of -3 corresponds to approximately 0.13% (or about 0.0013 in probability) in the lower tail.
- A \( z \)-score of 3 corresponds to approximately 99.87% (or about 0.9987 in probability) in the upper tail.
To find the percentage of customer accounts within this range, we calculate:
\[
\text{Percentage} = \text{Upper} - \text{Lower} = 99.87\% - 0.13\% \approx 99.74\%
\]
Thus, approximately **99.74%** of customer accounts have payments made between **6 days and 48 days**.