5 Marks Questions

Question 15 Marks

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: $x^{2}=6 y$

Answer

The given equation of parabola is $x^{2}=6 y$ which is of the form $x^{2}=4 a y$
$\therefore 4 \mathrm{a}=6 \Rightarrow a=\frac{6}{4} \Rightarrow a=\frac{3}{2}$
$\therefore$ Coordinates of focus are $\left(0, \frac{3}{2}\right)$
Axis of parabola is $x=0$
Equation of the directrix is $y=\frac{-3}{2} \Rightarrow 2 y+3=0$
Length of latus rectum $=\frac{4 \times 3}{2}=6$

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Question 25 Marks

The following table gives the number of finished articles turned out per day by different number of workers in a factory. Find the standard deviation of the daily output of finished articles.

Number of articles:	18	19	20	21	22	23	24	25	26	27
No. of workers:	3	7	11	14	18	17	13	8	5	4

Answer

For Calculation of Standard Deviation we prepare the following table.

$\text{x}$	$\text{f}$	$d_i=x_i-23$	$d_1^2$	$f _{ i } d _{ i }$	$f_i d_i^2$
18	3	-5	25	-15	75
19	7	-4	16	-28	112
20	11	-3	9	-33	99
21	14	-2	4	-28	56
22	18	-1	1	-18	18
23	17	0	0	0	0
24	13	1	1	13	13
25	8	2	4	16	32
26	5	3	9	15	45
27	4	4	16	16	64
	$N=\sum f_i=100$			$\sum f_i d_i=-62$	$\sum f_i d_i^2=514$

Clearly, $\mathrm{N}=100, \sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}=-62$ and $\sum f_{i} d_{i}^{2}=514$
$\therefore \sigma^{2}=\frac{1}{N}\left(\Sigma f_{i} d_{i}^{2}\right)-\left(\frac{1}{N} \Sigma f_{i} d_{i}\right)^{2}=\frac{514}{100}-\left(-\frac{62}{100}\right)^{2}=\frac{47556}{10000}$
Hence, $\sigma=\sqrt{\frac{47556}{10000}}=\frac{218.07}{100}=2.1807$

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Question 35 Marks

While calculating the mean and variance of 10 readings, a student wrongly uses the reading 52 for the correct reading 25 . He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and variance.

Answer

Number of observations, $n=10$
Mean,
$\bar{x}=45$ Variance,
$\sigma^{2}=16$
Now,
Incorrect mean,
$\bar{x}=45$
$\Rightarrow \frac{\text { Incorrect } \sum x_{i}}{10}=45$
$\Rightarrow$ Incorrect $\sum \mathrm{x}_{\mathrm{i}}=450$
$\therefore$ Correct $\sum \mathrm{x}_{\mathrm{i}}=450-52+25=423$
$\Rightarrow$ Correct mean $=\frac{\text { Correct } \sum x_{i}}{10}=\frac{423}{10}=42.3$
Incorrect variance,
$\sigma^{2}=16$
$\Rightarrow 16=\frac{\text { Incorrect } \sum x_{i}^{2}}{10}-(45)^{2}$
$\Rightarrow$ Incorrect $\sum x_{i}^{2}=10(16+2025)=20410$
$\therefore$ Correct $\sum x_{i}^{2}=20410-(52)^{2}+(25)^{2}=20410-2704+625=18331$
Now,
Correct variance $=\frac{18331}{10}-(42.3)^{2}=1833.1-1789.29=43.81$

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Question 45 Marks

For what value of $\lambda$ is the function defined by $f(x)=\left\{\begin{array}{cl}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{array}\right.$ continuous at $\mathrm{x}=0$?

Answer

$\underset{x=0}{L H L}=\underset{h\rightarrow0}{\text{lim}~} f(0-h)$
$=\underset{h\rightarrow0}{\text{lim}~} \lambda\left(h^{2}+2 h\right)=0$
$\underset{x=0}{RHL}=\underset{h\rightarrow0}{\text{lim}~} f(0+h)=\underset{h\rightarrow0}{\text{lim}~}(4 h+1)=1$
As for no value of $\lambda, \underset{x=0}{\mathrm{LHL}}=\underset{x=0}{\mathrm{RHL}},$
$\therefore$ function is not continuous at $\mathrm{x}=0$, for any value of $\lambda$.

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Question 55 Marks

In a bolt factory, three machines, A, B, C, manufacture $25 \%, 35 \%$ and $40 \%$ of the total production respectively. Of their respective outputs, $5 \%, 4 \%$ and $2 \%$ are defective. A bolt is drawn at random from the total product and it is found to be defective. Find the probability that it was manufactured by the machine C.

Answer

Let $E_{1}, E_{2}$ and $E_{3}$ be the events of drawing a bolt produced by machine $A, B$ and $C$ respectively. Therefore, we have, $\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{25}{100}=\frac{1}{4}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{35}{100}=\frac{7}{20}$, and $\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{40}{100}=\frac{2}{5}$
Let E be the event of drawing a defective bolt. Therefore,
$P\left(\frac{E}{E_{1}}\right)=$ probability of drawing a defective bolt, given that it is produced by the machine $\mathrm{A}=\frac{5}{100}=\frac{1}{20}$
$P\left(\frac{E}{E_{2}}\right)=$ probability of drawing a defective bolt, given that it is produced by the machine $\mathrm{B}=\frac{4}{100}=\frac{1}{25}$
$P\left(\frac{E}{E_{3}}\right)=$ probability of drawing a defective bolt, given that it is produced by the machine $\mathrm{C}=\frac{2}{100}=\frac{1}{50}$
Therefore, we have,
Probability that the bolt drawn is manufactured by C, given that it is defective
$=P\left(\frac{E_{3}}{E}\right)$
$=\frac{P\left(\frac{E}{E_{3}}\right) \cdot P\left(E_{3}\right)}{P\left(\frac{E}{E_{1}}\right) \cdot P\left(E_{1}\right)+P\left(\frac{E}{E_{2}}\right) \cdot P\left(E_{2}\right)+P\left(\frac{E}{E_{3}}\right) \cdot P\left(E_{3}\right)}$ [by Bayes's theorem]
$\frac{\left(\frac{1}{50} \times \frac{2}{5}\right)}{\left(\frac{1}{20} \times \frac{1}{4}\right)+\left(\frac{1}{25} \times \frac{7}{20}\right)+\left(\frac{1}{50} \times \frac{2}{5}\right)}=\left(\frac{1}{125} \times \frac{2000}{69}\right)=\frac{16}{69}$
Hence, the required probability is $\frac{16}{69}$.

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Question 65 Marks

The probability of a student A passing an examination is $\frac{3}{5}$ and of student $B$ is $\frac{4}{5}$. Assuming that the two events A passes, B passes as independent. Find the probability of
i. both the students passing the examination
ii. only A passing the examination
iii. only one of them passing the examination
iv. none of them passing the examination

Answer

i. P(both the students passing the examination)
$=\frac{3}{5} \times \frac{4}{5}=\frac{12}{25}$
ii. P(only student A passing the examination)
$=\frac{3}{5} \times \frac{1}{5}=\frac{3}{25}$
iii. P (only one of them passing the examination)
$=\mathrm{P}(\mathrm{A}$ passes and B does not pass) or (A does not pass and B passes)
$=\frac{3}{5} \times \frac{1}{5}+\frac{2}{5} \times \frac{4}{5}=\frac{3+8}{25}=\frac{11}{25}$
iv. P (none of them passing the examination)
$=\mathrm{P}(\mathrm{A}$ does not pass and B does not pass)
$=\frac{2}{5} \times \frac{1}{5}=\frac{2}{25}$

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Applied Maths 5 Marks Questions

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