Data representation

This statement is checking whether an ogive is drawn by plotting cumulative totals using the class midpoints as the x-values.

In statistics, especially in Statistical Measures and Data Representation, different graphs follow different x-axis choices:

1. Frequency Polygons: Plot frequencies at the class midpoints.
2. Cumulative Frequency Curves (Ogives): Plot cumulative frequencies at the upper class boundaries.

This makes sense because a cumulative frequency tells you how many values are less than or equal to a particular cutoff, and that cutoff is reached at the end of each class interval (its upper boundary).

An ogive uses the upper class boundary, not the midpoint, so the statement does not follow the standard rule.

Because the statement wrongly says to use midpoints for an ogive, the correct choice is False.

We need to decide if an ogive uses class midpoints on the x-axis. The easiest way is to compare how interval data is handled in different statistical graphs.

Midpoints are the standard x-values for frequency polygons.

In a frequency polygon, each class’s frequency is placed at the center of the interval to show the overall shape of the distribution. If the statement were about frequency polygons, it would be correct.

A cumulative frequency curve (ogive) shows the running total of frequencies up to a given value.

So we plot each cumulative total at the upper class boundary (the right-hand end of the interval), not the midpoint. For example, if an interval is $10\le x<20$, the cumulative frequency at that point represents how many values are less than $20$.

Since midpoints belong to frequency polygons and ogives use upper boundaries, the statement is wrong.

The statement is false because cumulative frequencies are plotted against the upper class boundaries of the intervals.

The statement claims that when drawing a cumulative frequency curve (ogive), we should put cumulative frequencies above the class midpoints on the x-axis.

To judge this, we should use the meaning of cumulative frequency from Topic: Statistical Measures and Data Representation.

A cumulative frequency for a class interval is the total number of data values that are less than or equal to a particular endpoint.

For example, if we have a class $10\le x<20$ with a cumulative frequency of $15$, that means there are $15$ values that are strictly less than $20$.

Because the cumulative total is counting values below a cutoff, the correct plotting position is the upper class limit.

$\text{Cumulative Frequency}\to \text{Upper Class Limit}$

Using the midpoint would wrongly suggest the total is guaranteed to be reached by the middle of the class, which grouped data does not tell us. Therefore the upper limit is the accurate choice.

The statement says to use midpoints, but the definition shows we must use upper class limits for the “less than or equal to” meaning. So the True option must be rejected.

Since an ogive plots cumulative frequency at upper class boundaries rather than midpoints, the correct answer is False.

Start by focusing on the action taken in the graph. The newsletter shows several years of fundraising totals but deliberately leaves out the year that would make the results look worse (the year with a big drop).

That means the display is not complete; it is choosing only the “nice-looking” years so the trend seems more positive than it really is.

A rounding error happens when values are rounded (up or down) and this creates small inaccuracies. Here, no one is rounding the totals—they are removing a whole year from the graph. So rounding error does not describe this situation.

Inappropriate grouping is when data is put into categories or intervals in a way that misleads (for example, using strange bins that hide changes). In this graph, the categories are still years; the problem is not the grouping method but that one year has been left out entirely.

Sampling bias occurs when the data collected comes from a non-representative sample (so the collection process is flawed). In this case, the totals for all years exist, but one year is simply not shown. That’s a reporting choice, not a biased sample.

By definition, cherry-picking (sometimes called using incomplete evidence) is when someone selects only the data that supports their message and ignores data that would challenge it.

Because the newsletter hid the year with the sharp drop to make the fundraising look consistently strong, this is a clear example of cherry-picking.

The important detail is that the newsletter is purposely not showing one specific year—the year with poor results—so the overall trend appears better. We are looking for the term that describes this kind of selective reporting.

A rounding error is about changing numbers slightly by rounding (for example, rounding $15.8\%$ to $16\%$), which affects precision. That is different from deleting an entire year of data, so this option does not fit.

Inappropriate grouping happens when the way data is placed into groups (like intervals) misleads the reader. Here, the time categories are still by year; the issue is that a year is missing, not that the years are grouped in a tricky way.

Sampling bias is a problem from how data is gathered, when the sample doesn’t represent the whole population. In this scenario, the data for all years is available, but one year is intentionally not included in the display. That is not sampling bias.

Cherry-picking means presenting only the information that supports a claim while hiding information that weakens it. By showing the successful years and leaving out the bad year, the newsletter is selectively choosing data to create a misleading impression.

With grouped data in Topic: Statistical Measures and Data Representation, the exact data values inside each class interval aren’t listed. To approximate the mean, we pick one value that stands in for every data point in that interval.

From the class interval given, we identify its endpoints.

• Lower bound = $40$
• Upper bound = $60$

Using the Midpoint Calculation approach, we take the mean of the two class boundaries to represent the whole interval:

$\text{Midpoint}=\frac{\text{Lower Bound}+\text{Upper Bound}}{2}$

Substituting our values:

$\begin{array}{rl}\text{Midpoint}& =\frac{40+60}{2}\\ & =\frac{100}{2}\\ & =50\end{array}$

To make sure we choose correctly, we can explain why the other options don’t work:

• Option A: This corresponds to just the lower limit of the class.
• Option B: This is the frequency, which counts how many values are in the class, not the value that represents them.
• Option D: This corresponds to just the upper limit of the class.

The midpoint is $50$, matching Option C. This is the number you would multiply by the frequency ($12$) to estimate the total contribution of this interval to the overall sum.

To calculate a mean from a grouped frequency table, the individual data values are unknown. So we use one representative value for the interval, called the class mark (or midpoint).

Here is what we are given:

• The interval is $40\le x<60$
• The frequency is $12$

Options A and D are the two boundaries of the class interval.

If we used $40$, we’d be assuming every value in the interval is as small as possible, which would make the estimate too low. If we used $60$, we’d be assuming every value is as large as possible, which would make the estimate too high. So neither boundary is used as the representative value.

Option B is $12$, which is the frequency.

Frequency tells us the number of data points in the interval, not a typical value from the interval. In mean calculations, frequency is multiplied by the representative value—it is not the representative value itself.

After eliminating the other choices, Option C remains.

The value $50$ is the class mark (midpoint) for the interval, found by averaging the boundaries:

$\text{Class Mark}=\frac{40+60}{2}=50$

This is the standard estimate used to find the sum of values for a grouped interval.

In statistics, a resistant measure is a number that does not get pushed around much by extreme data values (outliers). If you include a number that is far bigger or far smaller than the rest, a resistant measure changes only a little.

To decide whether the mean is resistant, we look at how it is computed. The arithmetic mean ($\bar{x}$) is found by adding all values and dividing by the number of values ($n$):

$\bar{x}=\frac{\sum x}{n}$

Since every data value is part of the sum, any extremely large (or small) value directly changes the total, which can drag the mean noticeably upward or downward.

Consider a quick example:

1. Original set: $\{1,2,3\}$ $\text{Mean}=\frac{1+2+3}{3}=2$
2. Add an extreme value: $\{1,2,3,100\}$ $\text{Mean}=\frac{1+2+3+100}{4}=26.5$

Just one outlier changed the mean from $2$ to $26.5$. That shows the mean is very affected by extreme values, so it is not resistant.

The claim says the mean is resistant because outliers do not affect it much. Our reasoning and example show the opposite: the mean can shift a lot when an outlier appears. So the first option cannot be correct.

Because extreme values can change the mean substantially, the statement is false. A measure such as the median is resistant because it depends mainly on the position/order of the data rather than the exact size of every value.

In statistics, a resistant measure (also called a robust measure) is one that does not change much when an extreme value (an outlier) is present. To check the claim, we can disprove it using a numerical counterexample, a key idea in Statistical Measures and Data Representation.

Begin with three values: $1,2,$ and $3$. Compute the mean ($\bar{x}$) by summing and dividing by how many numbers there are:

$\bar{x}=\frac{1+2+3}{3}=\frac{6}{3}=2$

Now include a very large outlier, $100$, so the data become $\{1,2,3,100\}$. An outlier is a value that sits far away from the rest of the dataset.

Compute the mean again for the new set:

$\text{New }\bar{x}=\frac{1+2+3+100}{4}=\frac{106}{4}=26.5$

The mean moved from $2$ to $26.5$ because of just one extreme value. That is a huge shift, so the mean is clearly pulled toward outliers and therefore is not resistant.

Since the counterexample shows the mean changes a lot with an outlier, the original statement cannot be true.

• Option A is wrong because the mean is not resistant.
• Option B is right because the statement is false.

A measure is called resistant if extreme values (outliers) do not strongly influence it. The question is asking whether the mean behaves that way.

The mean (arithmetic average) is computed by adding all values and dividing by the total number of values:

$\bar{x}=\frac{{\sum }_{i=1}^n{x}_i}{n}$

Because the calculation uses every data value, even one unusually large or small number can pull the mean toward it. That makes the mean sensitive rather than resistant.

The median is resistant because it depends on where values fall in order, not on adding them all.

For example, in $\{2,3,4,5,6\}$ the median is $4$. If we add the outlier $100$, we get $\{2,3,4,5,6,100\}$ and the median changes only to $4.5$. The outlier’s size does not drastically affect the median.

Since the mean can be strongly influenced by an outlier due to its dependence on the total sum, the statement that it is resistant is incorrect.

Because the mean is not resistant (unlike the median), Option A cannot be correct.

The statement is false: the mean is sensitive to outliers. Therefore, the correct choice is Option B.