Statistics: questions with answers

The class mark of a class is its mid-point, $\dfrac{\text{lower limit} + \text{upper limit}}{2}$. For $60 - 80$ that is $\dfrac{ 60 + 80 }{2} = 70$. $60$ is the class's lower limit and $80$ its upper limit; neither is ever used as $x_i$. $20$ is the class size, the width of the interval, not a value inside it.

From $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$ we get $\sum f_i d_i = (\bar{x} - a) \times \sum f_i = (24 - 25) \times 20 = -20$. $480$ is $\bar{x} \times \sum f_i$, with $a$ never subtracted at all. $-1$ is the difference $\bar{x} - a$ with the total frequency left out, and $20$ takes the difference the other way round.

Dividing every deviation by $h$ shrank the whole column by a factor $h$, so $\bar{x} = a + h \times \dfrac{\sum f_i u_i}{\sum f_i} = 55 + 20 \times \dfrac{ 40 }{ 20 } = 55 + 20 \times 2 = 95$. $57$ leaves the class size out and adds $\bar{u}$ straight to $a$; $40$ is $h\bar{u}$ with the assumed mean never added back; $15$ subtracts the correction instead of adding it.

The modal class is the class of greatest frequency. Here the greatest frequency is $19$, so the modal class is $105 - 110$ and its upper limit is $110$. $19$ itself is a COUNT of observations, not a value of the variable, so it can never be a limit. $105$ is the class's other limit and $100$ is where the table begins.

Rearranging, $h = (\text{Mode} - l) \times \dfrac{2f_1 - f_0 - f_2}{f_1 - f_0} = 4 \times \dfrac{ 5 }{ 2 } = 10$. $4$ is only $\text{Mode} - l$, the correction term, reported as though it were the class size. $23$ is the greatest frequency, a count of observations rather than a width, and $5$ is the denominator $2f_1 - f_0 - f_2$ of the bracket.

Write the mode formula with $f_0$ as the unknown: $\text{Mode} - l = \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, so $14 = \dfrac{ 18 - f_0}{2 \times 18 - f_0 - 15 } \times 20$, which gives $f_0 = 11$. $18$ is the modal-class frequency $f_1$, not the preceding-class frequency $f_0$ asked for. $25$ adds the difference $f_1 - f_0$ instead of subtracting it, and $7$ is that difference itself rather than the frequency asked for.

The table accumulates DOWNWARDS from the total: the entry against $130$ counts everything at or above it, $13$. What lies below is the rest, $38 - 13 = 25$. $13$ itself is the count on the other side of the limit - what a 'less than' reading of the same table hands over. $11$ is the frequency of a single class rather than a cumulative count, and $38$ is the whole distribution.

$\dfrac{n}{2}$ counts observations from the START of the distribution, so it can only be compared with a running total - the CUMULATIVE frequency. A class's own frequency is just a single count, not a running total. That is why a class can have a huge frequency and still never reach $\dfrac{n}{2}$ on its own. The class mark and the class size are values of the variable, not counts of anything.

The cumulative frequency of a class is the total frequency up to and including it, which is $25$ here. $15$ is that class's own frequency, not its cumulative frequency. $10$ is the running total of the classes before it, one row too early. $39$ is the total frequency of the whole table, which is the cumulative frequency of the last class.

The median lies in $100 - 110$, so the preceding cumulative frequency is $cf = 3 + x$. The formula gives $cf = 17 - 2 = 15$, hence $x = 12$. Including the median class in $cf$ gives $x = 2$. Reporting $15$ treats the cumulative count as the frequency of one class. The remaining option totals the frequencies outside the unknown class.

Use $l + ((n/2-cf)/f)h$. Here $l = 60$, $n/2 = 14$, $cf = 12$, $f = 10$ and $h = 20$. Substitution gives the median $64$. Subtracting $22$ includes the median class in the cumulative count. Omitting $h$ leaves a class fraction unscaled. Dividing by $28$ uses the whole distribution instead of the median-class count.

The total is $n = 28$, so $\dfrac{n}{2} = 14$; the first cumulative frequency past that belongs to $80 - 100$, so the median class is $80 - 100$ with $l = 80$, $cf = 11$, $f = 10$ and $h = 20$. Then $\text{Median} = l + \dfrac{\frac{n}{2} - cf}{f} \times h = 80 + \dfrac{ 3 }{ 10 } \times 20 = 86$. $66$ uses $21$, the median class's OWN cumulative frequency, in place of $cf$ - a negative numerator is the sign of it. $80.3$ adds the fraction to $l$ without multiplying by the class size. $14$ is $\dfrac{n}{2}$ itself, a position in the ordered data and not a value of the variable.

With $15$ observations the median is the $8$th when in order. Only $3$ observations were changed, all already above the median. Making them larger leaves them above it, so the $8$th observation stays at $20$. $22$ adds the increase to the median directly. $26$ multiplies the increase by how many observations changed. $21$ shifts the median by one position.

The total goes up by $10 \times 2 = 20$, but the mean is that total divided by $10$, so the mean goes up by exactly $2$, to $26$. $44$ multiplies the shift by the number of observations - it is the change in the TOTAL, not in the mean. $24$ leaves the mean unchanged, which is what happens to the spread rather than to an average, and $34$ adds the count instead of the shift.

The empirical relation is $3\,\text{Median} \approx \text{Mode} + 2\,\text{Mean}$. Rearranging gives $\text{Mean} \approx (3\,\text{Median}-\text{Mode})/2 = 40$. The expression $(\text{Mode}+2\,\text{Median})/3$ interchanges mean and median in the relation. Averaging the two given values and omitting division by two give the other distractors.