[8 marks Questions]

Question 18 Marks

Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Prices in city A	20	22	19	23	16
Prices in city B	10	20	18	12	15

Answer

Coefficients of variation of prices in city A.
Arrange in ascending order we get, 16, 19, 20, 22, 23
Assumed mean = 20

$x_i$	$d_i = x_i − A$	$d_i^2$
1	− 4	16
19	− 1	1
20	0	0
22	2	4
23	3	9
$\sum x_{ i }=100$	$\sum d _{ i }=0$	$\sum d _{ i }^2= 3 0$

For city A
$
\begin{aligned}
& \text { Here } \sum x_{ i }=100, \sum d _{ i }=0, \sum d _{ i }^2=30 \\
& \bar{x}=\frac{\sum x_{ i }}{ n } \\
& =\frac{100}{5} \\
& \Rightarrow \bar{x}=20
\end{aligned}
$

Standard deviation $(\sigma)=\sqrt{\frac{\sum d _{ i }^2}{ n }-\left(\frac{\sum d _{ i }}{ n }\right)^2}$
$
\begin{aligned}
& =\sqrt{\frac{30}{5}-\left(\frac{0}{5}\right)^2} \\
& =\sqrt{6}
\end{aligned}
$
$(\sigma)=2.45$
Coefficient of variation $=\frac{\sigma}{\bar{x}} \times 100$
$
=\frac{2.45}{20} \times 100
$
= 12.25%
Coefficient of variation = 12.25%
Coefficients of variation of prices in city B.
Arrange in ascending order we get, 10, 12, 15, 18, 20
Assumed mean = 15

$x_i$	$d_i = x_i − A$	$d_i^2$
10	− 5	25
12	− 3	9
15	0	0
18	3	9
20	5	25
$\sum x_{ i }=75$	$\sum d_{ i }=0$	$\sum d_{ i }^2=68$

For city B
$
\begin{aligned}
& \text { Here } \sum x_{ i }=75, \sum d _{ i }=0, \sum d _{ i }^2=68 \\
& \bar{x}=\frac{\sum x_{ i }}{ n }=\frac{75}{5}=15
\end{aligned}
$
$
\text { Standard deviation }(\sigma)=\sqrt{\frac{\sum d _{ i }^2}{ n }-\left(\frac{\sum d _{ i }}{ n }\right)^2}
$
$
\begin{aligned}
& =\sqrt{\frac{68}{5}-\left(\frac{0}{5}\right)^2} \\
& =\sqrt{13.6} \\
& =3.69 \\
& \text { Coefficient of variation }=\frac{\sigma}{\bar{x}} \times 100 \\
& =\frac{3.69}{15} \times 100 \\
& =24.6 \%
\end{aligned}
$
Prices in city $A$ is more stable ...(since $12.25 \%<24.6 \%)$

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Question 28 Marks

The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50 . Compute the missing frequencies $f_1$ and $f_2$.

Class Interval	0 − 20	20 − 40	40 − 60	60 − 80	80 − 100	100 − 120
Frequency	5	$f_1$	10	$f_2$	7	8

Answer

Arithmetic Mean $(\bar{x})=62.8$
Sum of all the frequencies $\left(\sum f_{ i }\right)=50$
Let the missing frequencies be $f_1$ and $f_2$

ss interval	frequency $f_i$	mid value $x_i$	$f_ix_i$
0 − 20	5	10	0
20 − 40	$f_1$	30	30 $f_1$
40 − 60	10	50	500
60 − 80	$f_2$	70	70 $f_2$
80 − 100	7	90	630
100 − 120	8	110	880
	$\sum f_{ i }=30+f_1+f_2$		$\sum f_{ i } x_{ i }=2060+30 f_1+70 f_2$

$
\begin{aligned}
& 30+f_1+f_2=50 \\
& f_1+f_2=20 \quad \ldots(1) \\
& \text { Mean }=\frac{\sum f_{ i } x_{ i }}{\sum f_{ i }} \\
& 62.8=\frac{2060+30 f_1+70 f_2}{50} \\
& 3140=2060+30 f_1+70 f_2 \\
& \therefore 30 f_1+70 f_2=3140-2060 \\
& 30 f_1+70 f_2=1080 \\
& (\div \text { by } 10) \Rightarrow 3 f_1+7 f_2=108 \quad \ldots(2) \\
& (1) \times 3 \Rightarrow 3 f_1+3 f_2=60 \quad \text {...(3) } \\
& (2) \times 3 \underline{\Rightarrow 3 f_1+7 f_2=108 \quad \ldots(4)} \\
& (3)-(4) {\Rightarrow}-4 f_2=-48 \\
& f_2=\frac{48}{4} \Rightarrow f_2=12 \\
&
\end{aligned}
$

Substitute of the value of $f_2$ in (1)
$
f_1+12=20
$
⇒ $f_1$ = 20 – 12 = 8
The Missing frequency is 8 and 12.

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Question 38 Marks

In a class of 35, students are numbered from 1 to 35. The ratio of boys to girls is 4 : 3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.

Answer

$
\begin{aligned}
& \text { Sample space }(S)=\{1,2,3, \ldots, 35\} \\
& n(S)=35
\end{aligned}
$

Total number of students $=35$
$
\begin{aligned}
& \text { Number of boys }=\frac{4}{7} \times 35 \\
& =20 \ldots[\text { Boys Numbers }=\{1,2,3, \ldots, 20\}]
\end{aligned}
$

Number of girls $=\frac{3}{7} \times 35$
$
=15 \ldots[\text { Girls Numbers }=\{21,22, \ldots, 35\}]
$

Let $A$ be the event of getting a boy role number with prime number
$
\begin{aligned}
& A=\{2,3,5,7,11,13,17,19\} \\
& n(A)=8 \\
& P(A)=\frac{n(A)}{n(S)}=\frac{8}{35}
\end{aligned}
$

Let $B$ be the event of getting girls roll number with composite number.
$
\begin{aligned}
& B=\{21,22,24,25,26,27,28,30,32,33,34,35\} \\
& n(B)=12 \\
& P(B)=\frac{n(B)}{n(S)}=\frac{12}{35} \\
& \text { Let } C \text { be the event of getting an even roll number. } \\
& C=\{2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34\} \\
& n(C)=17 \\
& P(C)=\frac{n(C)}{n(S)}=\frac{17}{35} \\
& n(A \cap B)=0 \\
& \Rightarrow P(A \cap B)=0 \\
& (B \cap C)=\{22,24,26,28,30,32,34\} \\
& n(B \cap C)=7 \\
& P(B \cap C)=\frac{n(B \cap C)}{n(S)}=\frac{7}{35}
\end{aligned}
$
$\begin{aligned} & (A \cap C)=\{2\} \\ & n(A \cap C)=1 \\ & P(A \cap C)=\frac{n(A \cap C)}{n(S)}=\frac{1}{35} \\ & A \cap B \cap C=\{\} \\ & n(A \cap B \cap C)=0 \\ & P(A \cap B \cap C)=0 \\ & P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(A \cap C)+P(A \cap B \cap C) \\ & =\frac{8}{35}+\frac{12}{35}+\frac{17}{35}-0-\frac{7}{35}-\frac{1}{35}+0 \\ & =\frac{8}{35}+\frac{12}{35}+\frac{17}{35}-\frac{8}{35} \\ & =\frac{8+12+17-8}{35} \\ & =\frac{29}{35}\end{aligned}$
Probability of getting roll number is $\frac{29}{35}$

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Question 48 Marks

A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads

Answer

$
\begin{aligned}
& \text { Sample space }=\{ HHH , HHT , HTH , HTT , THH , THT , TTH , TTT \} \\
& n ( S )=8
\end{aligned}
$

Let $A$ be the event of getting exactly two heads.
$
\begin{aligned}
& A=\{H H T, H T H, T H H\} \\
& n(A)=3 \\
& P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}
\end{aligned}
$

Let $B$ be the event of getting atleast one tail
$
\begin{aligned}
& B =\{ HHT , HTH , HTT , THH , THT , TTH , TTT \} \\
& n ( B )=7
\end{aligned}
$
$
P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}
$

Let $C$ be the event of getting consecutively
$
C =\{ HHH , HHT , THH \}
$
$\begin{aligned} & n(C)=3 \\ & P(C)=\frac{n(C)}{n(S)}=\frac{3}{8} \\ & A \cap B=\{H H T, H T H, T H H\} \\ & n(A \cap B)=3 \\ & p(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{3}{8} \\ & B \cap C=\{H H T, T H H\} \\ & n(B \cap C)=2 \\ & P(B \cap C)=\frac{n(B \cap C)}{n(S)}=\frac{2}{8} \\ & A \cap C=\{H H T, T H H\} \\ & n(A \cap C)=2 \\ & P(A \cap C)=\frac{n(A \cap C)}{n(S)}=\frac{2}{8}\end{aligned}$

$
\begin{aligned}
& n (A \cap B \cap C)=2 \\
& P(A \cap B \cap C)=\frac{n(A \cap B \cap C)}{n(S)}=\frac{2}{8} \\
& P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(A \cap C)+P(A \cap B \cap C) \\
& =\frac{3}{8}+\frac{7}{8}+\frac{3}{8}-\frac{3}{8}-\frac{2}{8}-\frac{2}{8}+\frac{2}{8} \\
& =\frac{3}{8}+\frac{7}{8}-\frac{2}{8} \\
& =\frac{8}{8} \\
& =1
\end{aligned}
$
The probability is 1 .

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Question 58 Marks

The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.

Answer

Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8

$x_i$	$x_i^2$
2	4
4	16
10	100
12	144
14	196
a	$a^2$
b	$b^2$
$\sum x_1=42+a+b$	$\sum x_i^2=460+a^2+b^2$

$\begin{aligned} & \frac{2+4+10+12+14+ a + b }{7}=8 \\ & \Rightarrow \frac{42+ a + b }{7}=8 \\ & a + b +42=56 \\ & a + b =56-42 \\ & a + b =14 \ldots(1) \\ & \text { variance }=16 \\ & \text { variance }=\frac{\sum x_{ i }^2}{ n }-\left(\frac{\sum x_{ i }}{ n }\right)^2 \\ & 16=\frac{460+ a ^2+ b ^2}{7}-\left(8^2\right) \\ & \Rightarrow 16=\frac{460+ a ^2+ b ^2}{7}-64 \\ & 16+64=\frac{460+ a ^2+ b ^2}{7}\end{aligned}$
$
\begin{aligned}
& \Rightarrow 80=\frac{460+a^2+b^2}{7} \\
& 560-460=a^2+b^2 \\
& a^2+b^2=100 \\
& \Rightarrow(a+b)^2-2 a b=100 \quad \ldots\left[a^2+b^2=(a+b)^2-2 a b\right] \\
& 14^2-2 a b=100 \\
& \Rightarrow 196-2 a b=100 \quad \ldots[a+b=14(\text { from }(1)] \\
& 196-100=2 a b \\
& 96=2 a b \\
& \Rightarrow a b=\frac{96}{2}=48 \\
& \therefore b=\frac{48}{a} \ldots(2)
\end{aligned}
$
Substitute the value of $b=\frac{48}{a}$ in (1)
$
a+\frac{48}{a}=14
$
⇒ $a^2$ + 48 = 14a
$a^2$ − 14a + 48 = 0
⇒ (a − 6) (a − 8) = 0
a = 6 or 8

When $a=6$
$
\begin{aligned}
& b=\frac{48}{a} \\
& =\frac{48}{6} \\
& =8
\end{aligned}
$

When $a=8$
$
\begin{aligned}
& b=\frac{48}{a} \\
& =\frac{48}{8} \\
& =6
\end{aligned}
$

∴ Missing observation is 8 and 6 or 6 and 8

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MATHS [8 marks Questions]

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