3 Marks Question

Question 13 Marks

If the probability of success in a single trial is $0 . 0 1$, how many minimum number of Bernoulli trials must be performed in order that the probability of at least one success is $\frac{ 1 }{ 2 }$ or more? (Use $\log _{10} 2=0.3010$ and $\log _{10} 99=1.9956$ )

Answer

We have, $p=0.01=\frac{1}{100} \Rightarrow q=\frac{99}{100}$
Let number of Bernoulli trials be $n$.
Now, the binomial distribution formula is for any random variable $( x )$ is given by
$
P(X=x)={ }^n C_x\left(\frac{1}{100}\right)^x\left(\frac{99}{100}\right)^{n-x}
$
So, the probability of at least one success is
$
P(X \geq 1)=1-P(X=0)=1-{ }^n C_0\left(\frac{1}{100}\right)^0\left(\frac{99}{100}\right)^n=1-\left(\frac{99}{100}\right)^n
$

According to condition, $P(X \geq 1) \geq 0.5 \Rightarrow 1-\left(\frac{99}{100}\right)^n \geq 0.5 \Rightarrow\left(\frac{99}{100}\right)^n \leq 0.5$
$
\Rightarrow n \log _{10} \frac{99}{100} \leq \log _{10} 0.5 \Rightarrow n \geq \frac{\log _{10} 0.5}{\log _{10} 0.99} ; \quad\left(\text { as } \log _{10} 0.99<0\right)
$
$\left[\right.$ Using $\log _{10} 2=0.3010$ and $\left.\log _{10} 99=1.9956\right] \Rightarrow n \geq 68.409 \Rightarrow n=69[\because n \in N ]$.

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Question 23 Marks

An octagonal prism is a three-dimensional polyhedron bounded by two octagonal bases and eight rectangular side faces. It has 24edges and 16vertices.

The prism is rolled along the rectangular faces and number on the bottom face (touching the ground) is noted. Let Xdenotes the number obtained on the bottom face and the following table gives the probability distribution of X.

X:	1	2	3	4	5	6	7	8
P(X):	p	2p	2p	p	2p	$p^2$	$2 p^2$	$7 p^2+p$

On the above context, answer the following questions.
(i) Find the value of $p$.
(ii) Find the mean, $E ( x )$.

Answer

(i) We have, $\sum_{i=1}^8 P(X=i)=1$
$\begin{array}{l}\Rightarrow p+2 p+2 p+p+2 p+p^2+2 p^2+7 p^2+p=1 \\ \Rightarrow 10 p^2+9 p-1=0 \\ \Rightarrow(10 p-1)(p+1)=0 \\ \Rightarrow p \neq-1 \\ \therefore p=\frac{1}{10}\end{array}$
(ii)
Mean, $E(X)=\sum_{i=1}^8 i P(X=i)$
$\begin{array}{l}=1 \times p+2 \times p+3 \times 2 p+4 \times p+5 \times 2 p+6 \times p^2+7 \times 2 p^2+8 \times\left(7 p^2+p\right) \\ =33 p+76 p^2 \\ =\frac{33}{10}+\frac{76}{100}=\frac{203}{50}\end{array}$

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Question 33 Marks

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 25. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 14. If the profit on a necklace is ₹ 100 and that on a bracelet is ₹ 300, formulate an L.P.P. for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.

Answer

Let the number of necklaces manufactured be $x$, and the number of bracelets manufactured be $y$.
According to question,
$
\begin{array}{l}
x+y \leq 25 \text { and } \\
\frac{x}{2}+y \leq 14
\end{array}
$
The profit on one necklace is ₹ $1 0 0$ and the profit on one bracelet is ₹ $3 0 0$. Let the profit (the objective function) be $Z$, which has to be maximized. Therefore, required LPP is
$
\text { Maximize } Z=100 x+ 3 0 0 y
$
Subject to the constraints
$x+y \leq 25$
$\frac{x}{2}+y \leq 14$
$x, y \geq 0$

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Question 43 Marks

Mr Rohit invested ₹ 5000 in a fund at the beginning of year 2021 and by the end of year 2021 his investment was worth ₹ 9000. Next year market crashed and he lost ₹ 3000 and ending up with ₹ 6000 at the end of year 2022. Next year i.e. 2023 he gained ₹ 4500 and ending up with ₹ 10500 at the end of the year. Find CAGR (Compounded Annual Growth Rate) of his investment.
$\left(\operatorname{Use}(2.1)^{1 / 3}=1.2805\right)$

Answer

Here, Initial investment value (IV) =₹ 5000
Final investment value (FV) =₹ 10500
No of period $(n)=3$ (starting from 2021 to 2023)
$
\begin{aligned}
\Rightarrow r & =\left(\frac{F V}{I V}\right)^{\frac{1}{n}}-1=\left(\frac{10500}{5000}\right)^{\frac{1}{3}}-1 \\
& =1.2805-1=0.2805
\end{aligned}
$
$
C A G R=28.05 \%
$

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Question 53 Marks

A traffic engineer records the number of bicycle riders that use a particular cycle track. He records that an average of 3.2 bicycle riders use the cycle track every hour. Given that the number of bicycles that use the cycle track follow a Poisson distribution, what is the probability that 2 or less bicycle riders will use the cycle track within an hour? Also find the mean expectation and variance for the random variable.
(Given $e^{-3.2}=0.041$ )

Answer

Given, mean $=\lambda=3.2$
Let $X$ be the number of bicycle riders which use the cycle track.
Required probability $=P(X \leq 2)=P(X=0)+P(X=1)+P(X=2)$
$\begin{array}{l}
=\frac{e^{-3.2}(3.2)^0}{0!}+\frac{e^{-3.2}(3.2)^1}{1!}+\frac{e^{-3.2}(3.2)^2}{2!} \\
=e^{-3.2}(1+3.2+5.12) \\
=0.041 \times 9.32=0.618\end{array}$
Also, mean expectation $=$ variance of $X=\lambda=3.2$

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Question 63 Marks

The manufacturer of electrical items makes bulbs and claims that these bulbs have a mean life of 25 months. The life in months of a random sample of 6 such bulbs are given to be 24, 26, 30, 20, 20 and 18. Test the validity of the manufacturer’s claim at 1% level of significance.
[Given $t_5(0.01)=4.032$ ]

Answer

Here, population mean $(\mu)=25$
Sample mean $(\bar{x})=\frac{\sum x_i}{n}=\frac{138}{6}=23$
Sample size $(n)=6$
Consider, Null hypothesis $H _{ 0 }$ : There is no significant difference between the sample mean and the population mean i.e., $\left(\mu_1=\mu_2\right)$.
Alternate hypothesis $H _\alpha$ : There is no significant difference between the sample mean and the population mean i.e., $\left(\mu_1 \neq \mu_2\right)$.
Values of $\left(x_i-\bar{x}\right)^2$ are $1,9,49,9,9$ and 25
$
\therefore s=\sqrt{\frac{102}{5}}=4.52
$
Now, $t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{23-25}{\frac{4.52}{\sqrt{6}}}$
=-1.09
$\Rightarrow|t|=1.09$
Since, calculated value $|t|=10.763<$ tabulated value $t_5(0.01)=4.132$
So, the null hypothesis is accepted.
Hence, the manufacturer's claim is valid at $1 \%$ level of significance.

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Question 73 Marks

Two batches of the same product are tested for their mean life. Assuming that, the lives of the product follow a normal distribution with an unknown variance; test the hypothesis that the mean life is the same for both the branches, given the following information:

Batch	Sample Size	Mean life (in hours)	Standard Deviation (in hours)
Batch I	10	750	12
Batch II	8	820	14

$\left[\right.$ Given $\sqrt{4.4444}=2.1081$ and $\left.t_{16}(0.05)=2.120\right]$

Answer

Given,
$
n_1=10, n_2=8, \overline{X_1}=750, \overline{X_2}=820, s_1=12 \& s_2=14
$
Consider, Null hypothesis $H _0$ : Mean life is same for both the batches i.e., $\left(\mu_1=\mu_2\right)$.
Alternate hypothesis $H _\alpha$ : Two batches have different mean lives i.e., $\left(\mu_1 \neq \mu_2\right)$.
Test Statistics,
$
\begin{array}{l}
t=\frac{\overline{x_1}-\overline{x_2}}{S} \times \sqrt{\frac{n_1 n_2}{n_1+n_2}}, \\
\text { Where } S=\sqrt{\frac{\left(n_1-1\right) s_1^2+\left(n_2-1\right) s_2^2}{n_1+n_2-2}} \\
\Rightarrow \quad S=\sqrt{\frac{9 \times 144+7 \times 196}{10+8-2}}
\end{array}
$
$
\begin{aligned}
& =\sqrt{\frac{2668}{16}}=12.91 \\
\therefore t & =\frac{750-820}{12.91} \times \sqrt{\frac{10 \times 8}{10+8}} \\
& =\frac{-70}{12.91} \times 2.1081 \\
& =-11.430
\end{aligned}
$
Since, calculated value $| t |= 1 1 . 4 3 0 >$ tabulated value $t _{ 1 6 }( 0 . 0 5 )= 2 . 1 2 0$
So, rejected the null hypothesis at $5 \%$ level of significance.
Hence, the mean life for both the batches is not the same.

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Question 83 Marks

Find the remainder when 5⁶¹ is divided by 7.

Answer

$
\begin{array}{l}
5 \equiv 5(\bmod 7) \\
\Rightarrow 5^2 \equiv 25(\bmod 7) \\
\Rightarrow 5^2 \equiv 4(\bmod 7) \\
\Rightarrow 5^4 \equiv 4^2(\bmod 7) \\
\Rightarrow 5^4 \equiv 2(\bmod 7) \\
\Rightarrow 5^{20} \equiv 32(\bmod 7) \\
\Rightarrow 5^{20} \equiv 4(\bmod 7) \\
\Rightarrow 5^{60} \equiv 1(\bmod 7) \\
\Rightarrow 5^{61} \equiv 5(\bmod 7)
\end{array}
$
Hence, the remainder when $5^{61}$ is divided by 7 is 5

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Applied Maths 3 Marks Question

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