Exercise-13.3, Class 10th, Maths, Chapter 13, NCERT

Q1

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units) — Number of consumers
65–85 : 4
85–105 : 5
105–125 : 13
125–145 : 20
145–165 : 14
165–185 : 8
185–205 : 4

Solution

1. Compute class-marks $x$ and $f$:

2. Mean

$$\bar{x}=\frac{\mathrm{\Sigma }fx}{\mathrm{\Sigma }f}=\frac{9320}{68}=137.0588\approx \mathbf{137.06}\text{ units}$$

3. Median

$N=68$. $N\mathrm{/}2=\mathrm{34<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Cumulative\; frequencies:</span>}$

Median class: $125-145$.
$l=125,h=20,f=20,\text{ c.f. before}=22$

Median formula:

$$\text{Median}=l+\frac{\frac{N}{2}-\text{c.f. before}}{f}\times h=125+\frac{34-22}{20}\times 20$$

$$=125+12=\mathbf{137.0}\text{ units}$$

4. Mode

Modal class = class with largest frequency = $125-145$with $f_1=20$
$f_0=13$ (preceding), $f_2=14$(succeeding), $l=125,h=20$

$$\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h=125+\frac{20-13}{40-13-14}\times 20$$

$$=125+\frac{7}{13}\times 20=125+10.7692\approx \mathbf{135.77}\text{ units}$$

Comparison / Interpretation:
Mean $\approx 137.06$, Median $=137.0$, Mode $\approx 135.77$ — all three measures are very close, indicating a fairly symmetric distribution around about 136–137 units.

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Q2

If the median of the distribution given below is $28.5$, find the values of $x$ and $y$.

Class interval — Frequency
0–10 : 5
10–20 : $x$
20–30 : 20
30–40 : 15
40–50 : $y$
50–60 : 5
Total = 60

Solution

Median is $28.5$ → it lies in class $20-30$. Use median formula:

Here $N=60$, so $N\mathrm{/}2=30$. Median class: $20-30$ with $l=20,h=10,f=20$. c.f. before = frequency up to previous class = $5+x$.

Median:

$$28.5=20+\frac{30-(5+x)}{20}\times 10=20+\frac{25-x}{20}\times 10=20+\frac{25-x}{2}$$

So

$$28.5=20+\frac{25-x}{2}\Rightarrow \frac{25-x}{2}=8.5\Rightarrow 25-x=17\Rightarrow x=8.$$

Now total frequencies:

$$5+x+20+15+y+5=60$$

Substitute $x=8$:

$$5+8+20+15+y+5=60\Rightarrow 53+y=60\Rightarrow y=7.$$

Answer: $x=\mathbf{8},y=\mathbf{7}$

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Q3

A life insurance agent found the following data for distribution of ages of 100 policy holders (cumulative ‘below’ form). Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years) — Number of policy holders (Below …)
Below 20 : 2
Below 25 : 6
Below 30 : 24
Below 35 : 45
Below 40 : 78
Below 45 : 89
Below 50 : 92
Below 55 : 98
Below 60 :100

Solution

First convert to class frequencies (by differences):

Total $N=100$. Median position $=N\mathrm{/}2=50$. Cumulative frequencies:

Median class = $35-40$. Use: $l=35,h=5,f=33,\text{ c.f. before}=45$

$$\text{Median}=35+\frac{50-45}{33}\times 5=35+\frac{5}{33}\times 5$$

$$=35+\frac{25}{33}\approx 35+0.7576=\mathbf{35.76}\text{ years (approx)}$$

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Q4

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre; the data:

Length (mm) — Number of leaves
118–126 : 3
127–135 : 5
136–144 : 9
145–153 : 12
154–162 : 5
163–171 : 4
172–180 : 2

Find the median length of the leaves.
(Hint: convert to continuous classes: 117.5–126.5, 126.5–135.5, …, 171.5–180.5.)

Solution

Use continuous class limits (class-width $h=9$):

$N=40,N\mathrm{/}2=20.$ Median class: $144.5-153.5$
$l$ c.f. before $=17$

$$\text{Median}=l+\frac{\frac{N}{2}-\text{c.f. before}}{f}\times h=144.5+\frac{20-17}{12}\times 9$$

$$=144.5+\frac{3}{12}\times 9=144.5+2.25=\mathbf{146.75}\text{ mm}$$

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Q5

The following table gives the distribution of the life time of 400 neon lamps. Find the median life time.

Life time (hours) — Number of lamps
1500–2000 : 14
2000–2500 : 56
2500–3000 : 60
3000–3500 : 86
3500–4000 : 74
4000–4500 : 62
4500–5000 : 48

Solution

Total $N=400,N\mathrm{/}2=200$. Cumulative frequencies:

Median class = $3000-3500$$l=3000,h=500,f=86,\text{ c.f. before}=130$

$$\text{Median}=3000+\frac{200-130}{86}\times 500=3000+\frac{70}{86}\times 500$$

$$=3000+0.8139535\times 500$$

$$=3000+406.9767\approx \mathbf{3406.98}\text{ hours}$$

Answer can be rounded: $\mathbf{3407}\text{ hours (approx)}$

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Q6

100 surnames were picked; frequency distribution of number of letters:

Number of letters — Number of surnames
1–4 : 6
4–7 : 30
7–10 : 40
10–13 : 16
13–16 : 4
16–19 : 4

Determine the median number of letters and find the mean number of letters.

Solution

1. Median

$N=100,N\mathrm{/}2=50.$ Cumulative frequencies:

Use continuous limits with class-width $h=3$ (e.g. 0.5–4.5, 4.5–7.5, 7.5–10.5 …). Median class interval in continuous form = $7.5-10.5$ So $l=7.5,h=3,f=40,$ c.f. before = 36.

$$\text{Median}=7.5+\frac{50-36}{40}\times 3=7.5+\frac{14}{40}\times 3=7.5+1.05=\mathbf{8.55}\text{ letters (approx)}$$

2. Mean

Class-marks $x$: 2.5, 5.5, 8.5, 11.5, 14.5, 17.5
Frequencies $f$: 6, 30, 40, 16, 4, 4

Compute $f\cdot x$:

$$\bar{x}=\frac{\mathrm{\Sigma }fx}{\mathrm{\Sigma }f}=\frac{832}{100}=\mathbf{8.32}\text{ letters}$$

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