Exercise 13.3 (Revised) - Chapter 13 - Statistics - Ncert Solutions class 10 - Maths

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Chapter 13 Statistics NCERT Solutions Class 10 Maths: Clear, Concise Answers

Ex 13.3 Question 1.

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.

Answer.

For Median: 

$\text { Here, } \sum f_i=n=68 \text {, then } \frac{n}{2}=\frac{68}{2}=34 \text {, which lies in interval } 125-145 \text {. }$

$\therefore$ Median class $=125-145$
So, $l=125, n=68, f=20, c f=22$ and $h=20$
Now, Median $=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h$
$
\begin{aligned}
& =125+\left[\frac{\frac{68}{2}-22}{20}\right] \times 20 \\
& =125+\frac{34-22}{20} \times 20=125+12=137
\end{aligned}
$

For Mean:

From given data, Assume mean $(a)=135$, Width of the class $(h)=20$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}
$

$
=\frac{7}{68}=0.102
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=135+20$ (0.102)
$
=135+2.04=137.04
$

For Mode:
In the given data, maximum frequency is 20 and it corresponds to the class interval 125 145.
$\therefore$ Modal class $=125-145$
And $l=125, f_1=20, f_0=13, f_2=14$ and $h=20$
$
\begin{aligned}
& \because \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\
& =125+\left[\frac{20-13}{2(20)-13-14}\right] \times 20 \\
& =125+\frac{7}{40-27} \times 20 \\
& =125+\frac{140}{13} \\
& =125+10.76923 \\
& =125+10.77 \\
& =135.77
\end{aligned}
$

Hence, median, mean and mode of given data is 137 units, 137.04 units and 135.77 units.

Ex 13.3 Question 2.

$\text {If the median of the distribution given below is } 28.5 \text {, then find the values of } x \text { and } y \text {. }$

Answer

Here, $\sum f_i=n=60$, then $\frac{n}{2}=\frac{60}{2}=30$, also, median of the distribution is 28.5 , which lies in interval $20-30$.
$\therefore$ Median class $=20-30$
So, $l=20, n=60, f=20, c f=5+x$ and $h=10$
$\because 45+x+y=60$
$\Rightarrow x+y=15$
Now, Median $=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h$
$\Rightarrow 28.5=20+\left[\frac{30-(5+x)}{20}\right] \times 10$

$
\begin{aligned}
& \Rightarrow 28.5=20+\frac{30-5-x}{2} \\
& \Rightarrow 28.5=\frac{40+25-x}{2} \\
& \Rightarrow 2(28.5)=65-x \\
& \Rightarrow 57.0=65-x \\
& \Rightarrow x=65-57=8 \\
& \Rightarrow x=8
\end{aligned}
$

Putting the value of $x$ in eq. (i), we get,
$
\begin{aligned}
& 8+y=15 \\
& \Rightarrow y=7
\end{aligned}
$

Hence the value of $x$ and $y$ are 8 and 7 respectively.
Ex 13.3 Question 3.

A life insurance agent found the following data for distribution of ages of $\mathbf{1 0 0}$ policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.

Answer

Here, $\sum f_i=n=100$, then $\frac{n}{2}=\frac{100}{2}=50$, which lies in interval $35-40$.
$\therefore$ Median class $=35-40$
So, $l=35, n=100, f=33, c f=45$ and $h=5$
Now, Median $=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h$
$=35+\left[\frac{\frac{100}{2}-45}{33}\right] \times 5$
$=35+\frac{50-45}{33} \times 5$
$=35+\frac{25}{33}$
$=35+0.7575$

$
\begin{aligned}
& =35+0.76 \text { (approx.) } \\
& =35.76
\end{aligned}
$

Hence median age of given data is 35.76 years.
Ex 13.3 Question 4.

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter and data obtained is represented in the following table. Find the median length of the leaves.

Answer.

Since the frequency distribution is not continuous, so firstly we shall make it continuous.

$\text { Here, } \sum f_i=n=40 \text {, then } \frac{n}{2}=\frac{40}{2}=20 \text {, which lies in interval } 144.5-153.5 \text {. }$

$\therefore$ Median class $=144.5-153.5$
So, $l=144.5, n=40, f=12, c f=17$ and $h=9$
$
\begin{aligned}
& \text { Now, Median }=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h \\
& =144.5+\left[\frac{20-17}{12}\right] \times 9 \\
& =144.5+\frac{3 \times 9}{12} \\
& =144.5+2.25 \\
& =146.75
\end{aligned}
$

Hence median length of the leaves is $146.75 \mathrm{~mm}$.
Ex 13.3 Question 5.

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

(change the frequency of class interval 3000-3500 from 85 to 86 )
Answer.

Here, $\sum f_i=n=400$, then $\frac{n}{2}=\frac{400}{2}=200$, which lies in interval $3000-3500$.
$\therefore$ Median class $=3000-3500$
So, $l=3000, n=400, f=86, c f=130$ and $h=500$
Now, Median $=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h$
$=3000+\left[\frac{200-130}{86}\right] \times 500$
$=3000+\frac{70 \times 500}{86}$
$=3000+406.9767441$
$=3000+406.98$ (approx.)
$=3406.98$
Hence median life time of a lamp is 3406.98 hours.

Ex 13.3 Question 6.

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also find the modal size of the surnames.

Ans. For Median:

Here, $\sum f_i=n=100$, then $\frac{n}{2}=\frac{100}{2}=50$, which lies in interval 7 -10 .
$\therefore$ Median class $=7-10$
So, $l=7, n=100, f=40, c f=36$ and $h=3$
Now, Median $=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h$
$
=7+\left[\frac{50-36}{40}\right] \times 3
$

$
\begin{aligned}
& =7+\frac{14 \times 3}{40} \\
& =7+1.05 \\
& =8.05
\end{aligned}
$

For Mean:

From given data, Assume mean $(a)=8.5$, Width of the class $(h)=3$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{-6}{100}=0.06
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=8.5+3(-0.06)=8.5-0.18=8.32$
For Mode:
In the given data, maximum frequency is 40 and it corresponds to the class interval $7-10$.
$\therefore$ Modal class $=7-10$
And $l=7, f_1=40, f_0=30, f_2=16$ and $h=3$
$
\because \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h
$

$
\begin{aligned}
& =7+\left[\frac{40-30}{2(40)-30-16}\right] \times 3 \\
& =7+\frac{10}{80-46} \times 3 \\
& =7+\frac{30}{34} \\
& =7+0.88 \text { (approx.) } \\
& =7.88
\end{aligned}
$

Hence, median, mean and mode of given data is 8.05 letters, 8.32 letters and 7.88 letters respectively.
Ex 13.3 Question 7.

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer

Here, $\sum f_i=n=30$, then $\frac{n}{2}=\frac{30}{2}=15$, which lies in interval $55-60$.
$\therefore$ Median class $=55-60$
So, $l=55, n=30, f=6, c f=13$ and $h=5$
Now, Median $=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h$
$=55+\left[\frac{15-13}{6}\right] \times 5$
$=55+\frac{2 \times 5}{6}$
$=55+1.66666$
$=5+1.67$ (approx.)
$=56.67$
Hence median weight of the students are $56.67 \mathrm{~kg}$.