Graphical Representation (Histograms and Ogives) — Important Questions

Answer: (b) Adjacent with no gaps.

Since the class intervals are continuous, the rectangles of a histogram touch each other with no gaps between them.

Answer: (b) $\dfrac{n}{2}$.

The median corresponds to the $\dfrac{n}{2}$th value, so the ordinate $\dfrac{n}{2}$ is used on a less-than ogive.

Answer: (c) A rising (non-decreasing) curve.

Cumulative frequency never decreases as the variable increases, so the less-than ogive rises from left to right.

Answer: (b) $16$.

Interquartile range $=Q_3-Q_1=40-24=16$.

Answer: (a) The ogive shows cumulative frequency versus upper boundaries, and the median is read at the $\dfrac{n}{2}$ ordinate on that curve. R correctly explains A.

The less-than cumulative frequencies are the running totals up to each upper boundary:

• less than $10$: $3$
• less than $20$: $3+7=10$
• less than $30$: $10+12=22$
• less than $40$: $22+8=30$

Total number of observations $n=30$.

Here $n=80$.

• Median: at $\dfrac{n}{2}=\dfrac{80}{2}=40$.
• Lower quartile $Q_1$: at $\dfrac{n}{4}=\dfrac{80}{4}=20$.
• Upper quartile $Q_3$: at $\dfrac{3n}{4}=\dfrac{3\times80}{4}=60$.

Draw horizontal lines at cumulative frequencies $40$, $20$ and $60$, and read the corresponding values on the $x$-axis.

Plot wages (class intervals) on the $x$-axis and frequency on the $y$-axis; draw adjacent rectangles of the given heights.

The tallest rectangle is the modal class $140$–$160$ (frequency $16$). To locate the mode, join the top-left corner of the modal rectangle to the top-left corner of the next rectangle, and the top-right corner of the modal rectangle to the top-right corner of the previous rectangle. From the point of intersection of these two lines drop a perpendicular to the $x$-axis; its foot $M$ gives the mode (about Rs $150$).

Here $n=50$.

Plot the points $(10,4),(20,14),(30,30),(40,44),(50,50)$ and join them with a smooth freehand curve (a less-than ogive).

• Median: draw a horizontal line at cumulative frequency $\dfrac{n}{2}=\dfrac{50}{2}=25$; where it meets the curve, drop a perpendicular to the $x$-axis to read the median height.
• Lower quartile $Q_1$: use the ordinate $\dfrac{n}{4}=\dfrac{50}{4}=12.5$.
• Upper quartile $Q_3$: use the ordinate $\dfrac{3n}{4}=\dfrac{3\times50}{4}=37.5$.

So horizontal lines are drawn at cumulative frequencies $25$, $12.5$ and $37.5$, and the corresponding $x$-values read from the curve give the median, $Q_1$ and $Q_3$ respectively.

Total candidates $n=200$.

Number scoring less than the pass mark $35$ (i.e. who failed) $=60$.

Number who passed $=200-60=140.$

Percentage who passed $=\dfrac{140}{200}\times100=70\%.$

So $140$ candidates passed, which is $70\%$ of all candidates.

Cumulative frequency (less-than) table:

• less than $10$: $5$
• less than $20$: $15$
• less than $30$: $35$
• less than $40$: $65$
• less than $50$: $85$
• less than $60$: $95$
• less than $70$: $100$

Plot $(10,5),(20,15),(30,35),(40,65),(50,85),(60,95),(70,100)$ and join with a smooth curve.

Median ($n=100$): ordinate $\dfrac{n}{2}=50$. The horizontal line at cf $=50$ meets the curve at marks $\approx35$, so median $\approx35$.

Lower quartile $Q_1$: ordinate $\dfrac{n}{4}=25$ gives $Q_1\approx25$.

Upper quartile $Q_3$: ordinate $\dfrac{3n}{4}=75$ gives $Q_3\approx45$.

Thus median $\approx35$, $Q_1\approx25$, $Q_3\approx45$, and the interquartile range $\approx45-25=20$.

Cumulative frequency (less-than) table:

• less than $60$: $8$
• less than $70$: $20$
• less than $80$: $40$
• less than $90$: $64$
• less than $100$: $80$

Plot $(60,8),(70,20),(80,40),(90,64),(100,80)$ and join with a smooth curve to form the less-than ogive.

Median ($n=80$): use ordinate $\dfrac{n}{2}=40$. The line at cf $=40$ meets the curve at wage $\approx80$, so the median wage $\approx$ Rs $80$.

Workers earning more than Rs $85$: at wage $=85$ the curve gives a cumulative frequency of about $52$ (interpolating between $40$ at $80$ and $64$ at $90$: $40+\dfrac{85-80}{10}\times24=40+12=52$). So the number earning less than Rs $85$ is about $52$, and the number earning more than Rs $85$ is

$80-52=28\text{ workers (approximately)}.$

(i) $n=40$ (the highest cumulative frequency).

(ii) Median is read at $\dfrac{n}{2}=\dfrac{40}{2}=20$.

(iii) $Q_1$ at $\dfrac{n}{4}=\dfrac{40}{4}=10$ and $Q_3$ at $\dfrac{3n}{4}=\dfrac{3\times40}{4}=30$.

(iv) From the table, the number of students scoring less than $30$ is the cumulative frequency at $30$, which is $20$ students.