Graphical Representation (Histograms and Ogives) — ICSE Class 10 Maths 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) $\dfrac{n}{2}$n/2.

The median corresponds to the $\dfrac{n}{2}$n/2th value, so the ordinate $\dfrac{n}{2}$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$16.

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

Answer: (a) The ogive shows cumulative frequency versus upper boundaries, and the median is read at the $\dfrac{n}{2}$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$10: $3$3
• less than $20$20: $3+7=10$3+7=10
• less than $30$30: $10+12=22$10+12=22
• less than $40$40: $22+8=30$22+8=30

Total number of observations $n=30$n=30.

Here $n=80$n=80.

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

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

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

The tallest rectangle is the modal class $140$140–$160$160 (frequency $16$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 $x$x-axis; its foot $M$M gives the mode (about Rs $150$150).

Here $n=50$n=50.

Plot the points $(10,4),(20,14),(30,30),(40,44),(50,50)$(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 $\dfrac{n}{2}=\dfrac{50}{2}=25$n/2=50/2=25; where it meets the curve, drop a perpendicular to the $x$x-axis to read the median height.
• Lower quartile $Q_1$Q_1: use the ordinate $\dfrac{n}{4}=\dfrac{50}{4}=12.5$n/4=50/4=12.5.
• Upper quartile $Q_3$Q_3: use the ordinate $\dfrac{3n}{4}=\dfrac{3\times50}{4}=37.5$3n/4=3×50/4=37.5.

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

Total candidates $n=200$n=200.

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

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

Percentage who passed $=\dfrac{140}{200}\times100=70\%.$=140/200×100=70\%.

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

Cumulative frequency (less-than) table:

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

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

Median ($n=100$n=100): ordinate $\dfrac{n}{2}=50$n/2=50. The horizontal line at cf $=50$=50 meets the curve at marks $\approx35$35, so median $\approx35$35.

Lower quartile $Q_1$Q_1: ordinate $\dfrac{n}{4}=25$n/4=25 gives $Q_1\approx25$Q_125.

Upper quartile $Q_3$Q_3: ordinate $\dfrac{3n}{4}=75$3n/4=75 gives $Q_3\approx45$Q_345.

Thus median $\approx35$35, $Q_1\approx25$Q_125, $Q_3\approx45$Q_345, and the interquartile range $\approx45-25=20$45-25=20.

Cumulative frequency (less-than) table:

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

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

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

Workers earning more than Rs $85$85: at wage $=85$=85 the curve gives a cumulative frequency of about $52$52 (interpolating between $40$40 at $80$80 and $64$64 at $90$90: $40+\dfrac{85-80}{10}\times24=40+12=52$40+85-80/10×24=40+12=52). So the number earning less than Rs $85$85 is about $52$52, and the number earning more than Rs $85$85 is

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

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

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

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

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