Thank you for visiting Suppose the pH of 7 10 in tap water was incorrectly recorded as 1 70 How does this affect the mean and the median What. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
To find out how the incorrect recording of the pH value affects the mean and median, let's break this down:
1. Understanding the Situation:
- We have a set of pH values from tap water that are correctly recorded as 7.1, 7.2, 7.3, 7.4, and 7.5.
- One of these values, 7.1, is incorrectly recorded as 1.7.
2. Calculating the Original Mean and Median:
- Mean: The average of a set of numbers.
[tex]\[
\text{Original Mean} = \frac{(7.1 + 7.2 + 7.3 + 7.4 + 7.5)}{5} = 7.3
\][/tex]
- Median: The middle value when the numbers are in order. Since we have an odd number of values, the median is the middle one.
[tex]\[
\text{Original Median} = 7.3
\][/tex]
3. Calculating the Incorrect Mean and Median:
- The new list of values is 1.7, 7.2, 7.3, 7.4, and 7.5.
- New Mean:
[tex]\[
\text{Incorrect Mean} = \frac{(1.7 + 7.2 + 7.3 + 7.4 + 7.5)}{5} \approx 6.220
\][/tex]
- New Median: The middle value in this ordered list is still 7.3.
4. Analysis:
- Effect on the Mean: The mean is affected because it takes into account all the values in the set. The incorrect value of 1.7, being significantly lower than the others, drags the average down from 7.3 to approximately 6.220.
- Effect on the Median: The median remains unchanged at 7.3. This shows a property of the median: it is resistant to extreme (outlier) values, unlike the mean.
In summary, the mean pH changes from 7.3 to approximately 6.220 due to the incorrect recording, while the median stays at 7.3, illustrating the median's robustness against outliers.
1. Understanding the Situation:
- We have a set of pH values from tap water that are correctly recorded as 7.1, 7.2, 7.3, 7.4, and 7.5.
- One of these values, 7.1, is incorrectly recorded as 1.7.
2. Calculating the Original Mean and Median:
- Mean: The average of a set of numbers.
[tex]\[
\text{Original Mean} = \frac{(7.1 + 7.2 + 7.3 + 7.4 + 7.5)}{5} = 7.3
\][/tex]
- Median: The middle value when the numbers are in order. Since we have an odd number of values, the median is the middle one.
[tex]\[
\text{Original Median} = 7.3
\][/tex]
3. Calculating the Incorrect Mean and Median:
- The new list of values is 1.7, 7.2, 7.3, 7.4, and 7.5.
- New Mean:
[tex]\[
\text{Incorrect Mean} = \frac{(1.7 + 7.2 + 7.3 + 7.4 + 7.5)}{5} \approx 6.220
\][/tex]
- New Median: The middle value in this ordered list is still 7.3.
4. Analysis:
- Effect on the Mean: The mean is affected because it takes into account all the values in the set. The incorrect value of 1.7, being significantly lower than the others, drags the average down from 7.3 to approximately 6.220.
- Effect on the Median: The median remains unchanged at 7.3. This shows a property of the median: it is resistant to extreme (outlier) values, unlike the mean.
In summary, the mean pH changes from 7.3 to approximately 6.220 due to the incorrect recording, while the median stays at 7.3, illustrating the median's robustness against outliers.
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Rewritten by : Jeany
