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
Answer
(b) Estimate the percentage of customer accounts for which the number of days is between 13 and 41
Answer
(c) Estimate the percentage of customer accounts for which the number of days is between 6 and 48
Answered by
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**.
Answered by
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**.
Answered by
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**.
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