[2 Mark Questions] - MATHS STD 10 Questions - Vidyadip

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[2 Mark Questions]

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30 questions · timed · auto-graded

Question 12 Marks

In a two children family, find the probability that there is at least one girl in a family

Answer

$
\begin{aligned}
& \text { Sample space }(S)=\{(\text { Boy, Boy) }(\text { Boy, Girl) (Girl, Boy) (Girl, Girl) }\} \\
& n(S)=4
\end{aligned}
$

Let $A$ be the event of getting atleast one Girl
$
\begin{aligned}
& A=\{(\text { Boy, Girl) (Girl, Boy) (Girl, Girl) }\} \\
& n(A)=3 \\
& P(A)=\frac{n(A)}{n(S)}=\frac{3}{4}
\end{aligned}
$

Probability of at least one girl in a family is $\frac{3}{4}$

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Question 22 Marks

If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data

Answer

Range of the data $(R)=20$
$L - S =20 \ldots(1)$

Coefficient of range $=0.2$
$
\begin{aligned}
& \text { Coefficient of range }=\frac{ L - S }{ L + S } \\
& 0.2=\frac{20}{ L + S } \\
& L + S =\frac{20}{0.2} \\
& =\frac{20 \times 10}{2} \\
& L + S =100 \quad \ldots(2) \\
& \underline{ L - S =20} \ldots(\text { (1) } \\
& \text { By adding (2) and }(1) \Rightarrow 2 L \quad=120 \\
& L \quad=\frac{120}{2}=60 \\
&
\end{aligned}
$
substituting the value of $L=60$ in (2)
$
60+S=100
$
S = 100 – 60 = 40
Largest value = 60
Smallest value = 40

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Question 32 Marks

The standard deviation of some temperature data in degree Celsius (°C) is 5. If the data were converted into degree Fahrenheit (°F) then what is the variance?

Answer

$
\begin{aligned}
& F ^{\circ}=\left( C ^{\circ} \times 1.8\right)+32 \\
& \sigma_{ C }=5^{\circ} C \\
& \sigma_{ F }=\left(1.8 \times 5^{\circ} C \right) \cdot 9^{\circ} F
\end{aligned}
$
Adding value to data doesn't affect standard deviation.
New variance $=\sigma_{ F }^2=81^{\circ} F$

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Question 42 Marks

A box contains cards numbered 3, 5, 7, 9, … 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.

Answer

Sample space $=\{3,5,7,9, \ldots 35,37\}$
$
n(S)=18
$

Let $A$ be the event of getting a multiple of 7
$
\begin{aligned}
& A=\{7,21,35\} \\
& n(A)=3 \\
& P(A)=\frac{n(A)}{n(S)}=\frac{3}{18}
\end{aligned}
$

Let $B$ be the event of getting a prime number
$
\begin{aligned}
& B=\{3,5,7,11,13,17,19,23,29,31,37\} \\
& n(B)=11
\end{aligned}
$
$
P(B)=\frac{n(B)}{n(S)}=\frac{11}{18}
$
$A \cap B=\{7\}$
$
n(A \cap B)=1
$

$
\begin{aligned}
& P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{18} \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& =\frac{3}{18}+\frac{11}{18}-\frac{1}{18} \\
& =\frac{3+11-1}{18} \\
& =\frac{13}{18}
\end{aligned}
$
Probability of getting a multiple of 7 or a prime number $=\frac{13}{18}$

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Question 52 Marks

From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being either a red king or a black queen

Answer

$
n(S)=52
$

Let $A$ be the event of getting a red king
$
\begin{aligned}
& n(A)=2 \\
& P(A)=\frac{n(A)}{n(S)}=\frac{2}{52}
\end{aligned}
$

Let B be the event of getting a black Queen king
$
\begin{aligned}
& n(B)=2 \\
& P(B)=\frac{n(B)}{n(S)}=\frac{2}{52}
\end{aligned}
$

It $A$ and $B$ are mutually exclusive
$
\begin{aligned}
& P(A \cup B)=P(A)+P(B) \\
& =\frac{2}{52}+\frac{2}{52} \\
& =\frac{4}{52}
\end{aligned}
$

$
=\frac{1}{13}
$
The required probability is $\frac{1}{13}$

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Question 62 Marks

The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B happen

Answer

Here P(A) = 0.5, P(B) = 0.3
P(A ∪ B) = P(A) + P(B) ...[A and B are mutually exclusive]
= 0.5 + 0.3
= 0.8
Probability of neither A nor B = [P(A ∪ B)']
= 1 – P(A ∪ B)
= 1 – 0.8
= 0.2

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Question 72 Marks

If A and B are two mutually exclusive events of a random experiment and P(not A) = 0.45, P(A ∪ B) = 0.65, then find P(B).

Answer

P(not A) = 0.45
1 – P(A) = 0.45
P(A) = 1 – 0.45 = 0.55
P(A ∪ B) = P(A) + P(B)
0.65 = 0.55 + P(B)
0.65 – 0.55 = P(B)
0.10 = P(B)
P(B) = 0.1

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Question 82 Marks

If $P(A)=\frac{2}{3}, P(B)=\frac{2}{5}, P(A \cup B)=\frac{1}{3}$, then find $P(A \cap B)$

Answer

$\begin{aligned} & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \frac{1}{3}=\frac{2}{3}+\frac{2}{5}-P(A \cap B) \\ & P(A \cap B)=\frac{2}{3}+\frac{2}{5}-\frac{1}{3} \\ & \Rightarrow \frac{10+6-5}{15} \\ & P(A \cap B)=\frac{11}{15}\end{aligned}$

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Question 92 Marks

The probability that a person will get an electrification contract is $\frac{3}{5}$ and the probability that he will not get plumbing contract is $\frac{5}{8}$. The probability of getting atleast one contract is $\frac{5}{7}$. What is the probability that he will get both?

Answer

Let $A$ and $B$ represent the event of getting electrification control and plumbing contract.
$
\begin{aligned}
& P(A)=\frac{3}{5} \\
& P(\overline B)=\frac{5}{8} \\
& 1-P(B)=\frac{5}{8} \\
& \therefore P(B)=1-\frac{5}{8}=\frac{8-5}{8} \\
& P(B)=\frac{3}{8} \\
& P(A \cup B)=\frac{5}{7} \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \frac{5}{7}=\frac{3}{5}+\frac{3}{8}-P(A \cap B) \\
& P(A \cap B)=\frac{3}{5}+\frac{3}{8}-\frac{5}{7}
\end{aligned}
$

$
\begin{aligned}
& =\frac{168+105-200}{280} \\
& =\frac{273-200}{280} \\
& =\frac{73}{280}
\end{aligned}
$
Probability of getting both the job is $\frac{73}{280}$

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Question 102 Marks

A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is neither white or black

Answer

Sample space $(S)=5+6+7+8$
$
n(S)=26
$

Let $A$ be the event of getting a white ball
$
n(A)=6
$
$
P(A)=\frac{n(A)}{n(S)}=\frac{6}{26}
$

Let $B$ be the event of getting a black ball
$
\begin{aligned}
& n(B)=8 \\
& P(B)=\frac{n(B)}{n(S)}=\frac{8}{26} \\
& P(A \cup B)=P(A)+P(B) \\
& =\frac{6}{26}+\frac{8}{26} \\
& =\frac{14}{26}
\end{aligned}
$

Probability of neither white or black $P(A \cup B)^{\prime}=1-P(A \cup B)$
$\begin{aligned} & =1-\frac{14}{26} \\ & =\frac{26-14}{26} \\ & =\frac{12}{26} \\ & =\frac{6}{13}\end{aligned}$

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Question 112 Marks

Two unbiased dice are rolled once. Find the probability of getting a doublet (equal numbers on both dice)

Answer

$
\begin{aligned}
& \text { Sample space }=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2), \\
& (3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1), \\
& (6,2),(6,3),(6,4),(6,5),(6,6)\} \\
& n(S)=36
\end{aligned}
$
Let $A$ be the event of getting doublet
$
\begin{aligned}
& A=\{(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)\} \\
& n(A)=6
\end{aligned}
$
$
P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}
$

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Question 122 Marks

Two unbiased dice are rolled once. Find the probability of getting the sum as a prime number

Answer

Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let C be the event of getting a sum is a prime number
C = {(1, 1) (1, 2) (1, 4) (1, 6) (2, 1) (2, 3) (2, 5) (3, 2), (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)}
n(C) = 15
$P ( C )=\frac{ n ( C )}{ n ( S )}=\frac{15}{36}=\frac{5}{12}$

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Question 132 Marks

A bag contains 12 blue balls and x red balls. If one ball is drawn at random what is the probability that it will be a red ball?

Answer

Sample space $=12+ x$
$
n(S)=x+12
$

Let $A$ be the event of getting a red ball
$
\begin{aligned}
& n(A)=x \\
& P(A)=\frac{n(A)}{n(S)} \\
& =\left(\frac{x}{x+12}\right)
\end{aligned}
$

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Question 142 Marks

At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that the first player wins a prize?

Answer

$22^2=484$
$31^2=961$
$23^2=529$
$32^2=1024$
$23,24,25,26,27,28,29,30,31$ has squares below $500 \times 1000$.
$P($ first player wins a prize $)=\frac{9}{1000}$

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Question 152 Marks

At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that the second player wins a prize if the first has won?

Answer

$\begin{aligned} & 22^2=484 \\ & 31^2=961 \\ & 23^2=529 \\ & 32^2=1024 \\ & 23,24,25,26,27,28,29,30,31 \text { has squares below } 500 \times 1000 \\ & P(\text { second player wins if first has won })=\frac{8}{999}\end{aligned}$

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Question 162 Marks

A coin is tossed thrice. What is the probability of getting two consecutive tails?

Answer

Sample space $=\{H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T\}$
$
n(S)=8
$

Let $A$ be the event of getting consecutive tails
$
\begin{aligned}
& A =\{ HTT , TTH , TTT \} \\
& n ( A )=3 \\
& P ( A )=\frac{ n ( A )}{ n ( S )}=\frac{3}{8}
\end{aligned}
$

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Question 172 Marks

If $A$ is an event of a random experiment such that $P(A): P(\bar{A})=17: 15$ and $n(S)=640$ then find $P (\overline{ A })$

Answer

$\begin{aligned} & P ( A ): P (\overline{ A })=17: 15 \\ & \frac{ P ( A )}{ P (\overline{ A })}=\frac{17}{15} \\ & \Rightarrow \frac{1- P (\overline{ A })}{ P (\overline{ A })}=\frac{17}{15} \\ & 17 P (\overline{ A })=15[1- P (\overline{ A })] \\ & =15-15 P (\overline{ A }) \\ & 17 P (\overline{ A })+15 P (\overline{ A })=15 \\ & \Rightarrow 32 P (\overline{ A })=15 \\ & P (\overline{ A })=\frac{15}{32}\end{aligned}$

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Question 182 Marks

If $A$ is an event of a random experiment such that $P(A): P(\bar{A})=17: 15$ and $n(S)=640$ then find $n(A)$

Answer

$\begin{aligned} & P (\overline{ A })=\frac{15}{32} \\ & 1- P ( A )=\frac{15}{32} \\ & \Rightarrow P ( A )=1-\frac{15}{32} \\ & =\frac{32-15}{32} \\ & =\frac{17}{32} \\ & P ( A )=\frac{ n ( A )}{ n ( S )}=\frac{17}{32} \\ & \Rightarrow \frac{ n ( A )}{640}=\frac{17}{32} \ldots[\text { Given } n ( S )=640] \\ & 32 n ( A )=17 \times 640 \\ & n ( A )=\frac{17 \times 640}{32} \\ & =17 \times 20\end{aligned}$
⇒ n(A) = 340

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Question 192 Marks

Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram)

Answer

Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

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Question 202 Marks

Write the sample space for tossing three coins using tree diagram

Answer

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

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Question 212 Marks

In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is $\frac{3}{8}$ then, find the number of defective bulbs

Answer

Let the number of defective bulbs be " $x$ "

Sample space $(S)=20+x$
$
n(S)=20+x
$

Let $A$ be the event of getting to be defective
$
\begin{aligned}
& n ( A )= x \\
& P ( A )=\frac{ n ( A )}{ n ( S )} \\
& \Rightarrow \frac{3}{8}=\frac{x}{20+x} \\
& \Rightarrow 8 x =3(20+ x )=(60+3 x ) \\
& \Rightarrow 8 x -3 x =60 \\
& \Rightarrow 5 x =60 \\
& \Rightarrow x =\frac{60}{5} \\
& \Rightarrow x =12
\end{aligned}
$
Number of defective bulbs = 12

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Question 222 Marks

The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below

Subject	Mean	SD
Mathematics	56	12
Science	65	14
Social Science	60	10

Which of the three subjects shows more consistent and which shows less consistent in marks?

Answer

(i) Mathematics:
Mean $(\bar{x})=56$
Standard deviation $(\sigma)=12$
Coefficient variation $\left( CV _1\right)=\frac{\sigma}{\bar{x}} \times 100$
$
\begin{aligned}
& =\frac{12}{56} \times 100 \\
& =\frac{1200}{56} \\
& =21.43
\end{aligned}
$

(ii) Science:
Mean $(\bar{x})=65$, Standard deviation $(\sigma)=14$
Coefficient variation $\left( CV _2\right)=\frac{14}{65} \times 100$
$
\begin{aligned}
& =\frac{1400}{65} \\
& =21.54
\end{aligned}
$

(iii) Social:
Mean $(\bar{x})=60$, Standard devaition $(\sigma)=10$
Coefficient variation $\left( CV _3\right)=\frac{10}{60} \times 100$
$
\begin{aligned}
& =\frac{1000}{60} \\
& =16.67
\end{aligned}
$
Science shows the more consistent
Social science shows the less consistent

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Question 232 Marks

If $n =5, \bar{x}=6, \sum x^2=765$, then calculate the coefficient of variation.

Answer

$\begin{aligned} & \text { Standard deviation }(\sigma)=\sqrt{\frac{\sum x^2}{ n }-\left(\frac{\sum x}{ n }\right)^2} \\ & =\sqrt{\frac{765}{5}-(6)^2} \\ & =\sqrt{153-36} \\ & =\sqrt{117} \\ & \sigma=10.816 \\ & \text { Coefficient of variation }=\frac{\sigma}{\bar{x}} \times 100 \% \\ & =\frac{10.816}{6} \times 100 \% \\ & =180.266 \% \\ & \text { Coefficient of variation }=180.27 \%\end{aligned}$

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Question 242 Marks

If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.

Answer

$
\begin{aligned}
& \text { Mean }(\bar{x})=15 \\
& \text { Coefficient of variation }=48 \\
& \frac{\sigma}{\bar{x}} \times 100=48 \\
& \Rightarrow \frac{\sigma}{15} \times 100=48 \\
& \sigma \times 100=48 \times 15 \\
& \sigma=\frac{48 \times 15}{100} \\
& =\frac{720}{100} \\
& =7.2
\end{aligned}
$
Standard deviation $(\sigma)=7.2$

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Question 252 Marks

The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.

Answer

Standard deviation $(\sigma)=1.2$
$
\begin{aligned}
& \text { Coefficient of variation }=25.6 \\
& \frac{\sigma}{\bar{x}} \times 100=25.6 \\
& \frac{1.2}{\bar{x}} \times 100=25.6 \\
& \Rightarrow 25.6 \times \bar{x}=1.2 \times 100 \\
& \bar{x}=\frac{120}{25.6} \\
& =\frac{120 \times 10}{256} \\
& =4.687 \\
& =4.69
\end{aligned}
$
Value of mean $=4.69$

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Question 262 Marks

If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation

Answer

The standard deviation of the data $=3.6$
Each value of the data is divided by 3
New standard deviation $=\frac{3.6}{3}=1.2$
New Variance $=(1.2)^2=1.44 \ldots\left[\therefore\right.$ Variance $\left.=(S . D)^2\right]$
New standard Deviation $=1.2$
New variance $=1.44$

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Question 272 Marks

Find the standard deviation of the first 21 natural numbers.

Answer

Here $n =21$

The standard deviation of the first ' $n$ ' natural numbers,
$
\begin{aligned}
& =\sqrt{\frac{ n ^2-1}{12}} \\
& 1,2,3,4, \ldots, 21=\sqrt{\frac{21^2-1}{12}} \\
& =\sqrt{\frac{441-1}{12}} \\
& =\sqrt{\frac{440}{12}} \\
& =\sqrt{36.666} \\
& =\sqrt{36.67} \\
& =6.055 \\
& =6.06
\end{aligned}
$
The standard deviation of the first 21 natural numbers $=6.06$

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Question 282 Marks

A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.

Answer

The remaining number of pages to be completed is 60 – 32, 60 – 35, 60 – 37, 60 – 30, 60 – 33, 60 – 36, 60 – 35 and 60 – 37
The pages to be completed are, 28, 25, 23, 30, 27, 24, 25, and 23
Arrange in ascending order we get, 23, 23, 24, 25, 25, 27, 28 and 30

x_(i)	x_(i)^(2)
23	529
23	529
24	576
25	625
25	625
27	729
28	784
30	900
sumx_(i)=205	sumx_(i)^(2)=5297

$
\begin{aligned}
& \text { Here } n =8, \sum x_{ i }=205, \sum x_{ i }^2=5297 \\
& \text { Standard deviation }(\sigma)=\sqrt{\frac{\sum x_{ i }^2}{ n }-\left(\frac{\sum x_{ i }}{ n }\right)^2} \\
& =\sqrt{\frac{5297}{8}-\left(\frac{205}{8}\right)^2} \\
& =\sqrt{662.13-\frac{42025}{64}} \\
& =\sqrt{662.13-656.64} \\
& =\sqrt{5.49} \\
& =2.34
\end{aligned}
$
Standard deviation $(\sigma)=2.34$

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Question 292 Marks

The time taken by 50 students to complete a 100-meter race are given below. Find its standard deviation

Time taken (sec)	8.5 − 9.5	9.5 − 10.5	10.5 − 11.5	11.5 − 12.5	12.5 − 13.5
Number of students	6	8	17	10	9

Answer

Assumed mean = 11

Time taken	Number of students $f_i$	mid value $x_i$	$d_i$ = $x_i$ − A = $x_i$ − 11	$f_id_i$	$f_id_i^2$
8.5 − 9.5	6	9	− 2	− 12	24
9.5 − 10.5	8	10	− 1	− 8	8
10.5 − 11.5	17	11	0	0	0
11.5 − 12.5	10	12	1	10	10
12.5 − 13.5	9	13	2	18	36
	$\sum f_{ i }=50$			$\sum f_{ i } d_{ i }=8$	$\sum f_{ i } d _{ i }^2=78$

$\begin{aligned} & \text { Here } N =50, \sum f_{ i } d _{ i }=8, \sum f_{ i } d _{ i }^2=78 \\ & \text { Standard deviation }(\sigma)=\sqrt{\frac{\sum f_{ i } d _{ i }^2}{ N }-\left(\frac{\sum f_{ i } d _{ i }}{ N }\right)^2} \\ & =\sqrt{\frac{78}{50}-\left(\frac{8}{50}\right)^2} \\ & =\sqrt{1.56-(0.16)^2} \\ & =\sqrt{1.56-0.025} \\ & =\sqrt{1.56-0.03} \\ & =\sqrt{1.53} \\ & =1.236 \\ & \therefore \text { Standard deviation }(\sigma)=1.24\end{aligned}$

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Question 302 Marks

The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation

Diameter (cm)	21 − 24	25 − 28	29 − 32	33 − 36	37 − 40	41 − 44
Number of plates	15	18	20	16	8	7

Answer

Assumed mean = 34.5

ameter (cm)	Number of plates $f_i$	mid value $x_i$	$d_i = x_i − A$ $= x_i − 34.5$	$f_id_i$	$f_id_i^2$
21 − 24	15	22.5	− 12	− 180	2160
25 − 28	18	26.5	− 8	− 144	1152
29 − 32	20	30.5	− 4	− 80	320
33 − 36	16	34.5	0	0	0
37 − 40	8	38.5	4	32	128
41 − 44	7	42.5	8	56	448
	$\sum f_{ i }=84$			$\sum f_{ i } d_{ i }=-316$	$\sum f_{ i } d _{ i }^2=4208$

Here $\sum f_{ i }=84, \sum f_{ i } d_{ i }=-316, \sum f_{ i } d _{ i }^2=4208$
Standard deviation $(\sigma)=\sqrt{\frac{\sum f_{ i } d _{ i }^2}{\sum f_{ i }}-\left(\frac{\sum f_{ i } d _{ i }}{\sum f_{ i }}\right)^2}$
$=\sqrt{\frac{4208}{84}-\left(\frac{316}{84}\right)}$
$=\sqrt{50.095-(3.76)^2}$
$=\sqrt{50.1-14.14}$
$=\sqrt{35.96}$
$=5.996$
$=6$
$\therefore$ Standard deviation $(\sigma)=6$

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