Exercise 13.2 (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 13.2 Question 1.

The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer.

For Mode: In the given data, maximum frequency is 23 and it corresponds to the class interval $35-45$.
$\therefore$ Modal class $=35-45$
And $l=35, f_1=23, f_0=21, f_2=14$ and $h=10$
$
\begin{aligned}
& \because \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\
& =35+\left[\frac{23-21}{2(23)-21-14}\right] \times 10 \\
& =35+\frac{2}{46-35} \times 10 \\
& =35+\frac{20}{11} \\
& =35+1.8
\end{aligned}
$

$
=36.8
$

For Mean:

From given data, Assume mean $(a)=30$, Width of the class $(h)=10$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{43}{80}=0.5375
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=30+10(0.5375)=30+5.375=35.37$
Hence mode of given data is 36.8 years and mean of the given data is 35.37 years.
Also, it is clear from above discussion that average age of a patient admitted in the hospital is 35.37 years and maximum number of patients admitted in the hospital are of age 36.8 years.
Ex 13.2 Question 2.

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components. 

Answer.

Given: Maximum frequency is 61 and it corresponds to the class interval $60-80$.
$\therefore$ Modal class $=60-80$
And $l=60, f_1=61, f_0=52, f_2=38$ 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 \\
& =60+\left[\frac{61-52}{2(61)-52-38}\right] \times 20 \\
& =60+\frac{9}{122-52-38} \times 20 \\
& =60+\frac{9}{32} \times 20 \\
& =60+5.625 \\
& =65.625
\end{aligned}
$

Hence modal lifetimes of the components is 65.625 hours.
Ex 13.2 Question 3.

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure:

Answer.

For Mode: Here, Maximum frequency is 40 and it corresponds to the class interval 1500 -2000 .
$\therefore$ Modal class $=1500-2000$
And $l=1500, f_1=40, f_0=24, f_2=33$ and $h=500$
$
\begin{aligned}
& \because \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\
& =1500+\left[\frac{40-24}{2(40)-24-33}\right] \times 500 \\
& =1500+\frac{16}{80-24-33} \times 500 \\
& =1500+\frac{8000}{23} \\
& =1500+347.83 \\
& =1847.83
\end{aligned}
$

For Mean:

From given data, Assume mean $(a)=2750$, Width of the class $(h)=500$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{-35}{200}=-0.175
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=2750+500(-0.175)=2750-87.50=2662.50$
Hence the modal monthly expenditure of family is Rs. 1847.83 and the mean monthly expenditure is Rs. 2662.50 .
Ex 13.2 Question 4.

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Ans. For Mode: Here, Maximum frequency is 10 and it corresponds to the class interval 30 35.
$\therefore$ Modal class $=30-35$
And $l=30, f_1=10, f_0=9, f_2=3$ and $h=5$
$
\because \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h=30+\left[\frac{10-9}{2(10)-9-3}\right] \times 5
$

$
=30+\frac{1}{20-12} \times 5=30+\frac{5}{8}=30+0.625=30.63 \text { (approx.) }
$

For Mean:

From given data, Assume mean $(a)=32.5$, Width of the class $(h)=5$
$
\therefore \bar{u}=\frac{\sum f_i u_i}{\sum f_i}=\frac{-23}{35}=-0.65
$

Using formula, Mean $(\bar{x})=a+h \bar{u}=32.5+5(-0.65)=32.5-3.25=29.25$ (approx.)
Hence mode and mean of given data is 30.63 and 29.25. Also from above discussion, it is clear that states/U.T. have students per teacher is 30.63 and on average, this ratio is 29.25 .
Ex 13.2 Question 5.

The given distribution shows the number of runs scored by some top batsmen of the world in one-day cricket matches:

Find mode of the data.
Answer.

In the given data, maximum frequency is 18 and it corresponds to the class interval $4000-5000$.
$\therefore$ Modal class $=4000-5000$
And $l=4000, f_1=18, f_0=4, f_2=9$ and $h=1000$
$
\begin{aligned}
& \because \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\
& =4000+\left[\frac{18-4}{2(18)-4-9}\right] \times 1000 \\
& =4000+\frac{14}{36-13} \times 1000 \\
& =4000+\frac{14000}{23} \\
& =4000+608.6956 \\
& =4608.7 \text { (approx.) }
\end{aligned}
$

Hence, mode of the given data is 4608.7 runs.

Ex 13.2 Question 6.

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below:

Find the mode of the data.
Answer.

In the given data, maximum frequency is 20 and it corresponds to the class interval 40 50.
$\therefore$ Modal class $=40-50$
And $l=40, f_1=20, f_0=12, f_2=11$ and $h=10$
$
\begin{aligned}
& \because \text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\
& =40+\left[\frac{20-12}{2(20)-12-11}\right] \times 10 \\
& =40+\frac{8}{40-23} \times 10 \\
& =40+\frac{80}{17}=40+4.70588=44.7 \text { (approx.) }
\end{aligned}
$

Hence, mode of the given data is 44.7 cars.