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Applied Maths Case study (4 Marks)

Model Paper 7 · CBSE STD 11 Science (CBSE - English Medium). 4 questions.

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Case study (4 Marks)

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Question 14 Marks
Read the text carefully and answer the questions
A shopkeeper sells three types of flower seeds $A _1, A_2$, and $A _3$ They are sold as a mixture, where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively.
(a) Calculate the probability of a randomly chosen seed to germinate.
(b) Calculate the probability that it is of the type A2 given that randomly chosen seed does not germinate.
(c) Calculate the probability that it will not germinate given that the seed is of type $A _1$.
Answer
A shopkeeper sells three types of flower seeds $A _1, A_2$, and $A _3$ They are sold as a mixture, where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively.
(i) We have: $A _1: A _2: A _3=4: 4: 2$
$P\left(A_1\right)=\frac{4}{10}, P\left(A_2\right)=\frac{4}{10}, P\left(A_3\right)=\frac{2}{10}$
where, $A_1, A_2$ and $A_3$ denote three types of flowers seed.
Let 'E' be the event that a seed germinates.
$\therefore P\left(\frac{E}{A_1}\right)=\frac{45}{100}, P\left(\frac{E}{A_2}\right)=\frac{60}{100} \& P\left(\frac{E}{A_2}\right)=\frac{35}{100}$
$P\left(\frac{\bar{E}}{A_1}\right)=\frac{55}{100}, P\left(\frac{\bar{E}}{A_2}\right)=\frac{40}{100} \& P\left(\frac{\bar{E}}{A_3}\right)=\frac{65}{100}$
$P(E)=P\left(A_1\right) \cdot P\left(\frac{E}{A_1}\right)+P\left(A_2\right) P\left(\frac{E}{A_2}\right)+P\left(A_3\right)\left(\frac{E}{A_3}\right)$
$\begin{array}{l}=\frac{4}{10} \cdot \frac{45}{100}+\frac{4}{10} \times \frac{60}{100}+\frac{2}{10} \cdot \frac{35}{100} \\ =\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=0.49\end{array}$
(ii)
$P\left(\frac{A_2}{\bar{E}}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{\bar{E}}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{\bar{E}}{A_1}\right)+P\left(A_2\right) \cdot P\left(\frac{\bar{E}}{A_2}\right)+P\left(A_3\right) \cdot\left(\frac{\bar{E}}{A_3}\right)}$
$\begin{array}{l}=\frac{\frac{4}{10} \cdot \frac{40}{100}}{\frac{4}{10} \cdot \frac{55}{100}+\frac{4}{10} \cdot \frac{40}{100}+\frac{2}{10} \cdot \frac{65}{100}} \\ =0.314\end{array}$
(iii) $P\left(\frac{E}{A_1}\right)=1-P\left(\frac{E}{A_1}\right)$
$=1-\frac{45}{100}$
$=\frac{55}{100}$
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Question 24 Marks
Answer
Vijay and Ramesh are playing cards. Total number of playing cards are 52 in numbers. Following events happen while playing the game of choosing a card out of 52 cards.
Image
(i) Favourable events are:
Total red cards = 26 (includes two kings)
Two black kings = total cards = 28
Total cards left (favourable) = 52 - 28 = 24
Required probability $=\frac{24}{52}=\frac{4}{13}$
(ii) Total number of aces = 4
Required probability $=\frac{4}{52}=\frac{1}{13}$
(iii) Total number of red cards = 26
Required probability $=\frac{26}{52}=\frac{1}{2}$
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Question 34 Marks
Read the text carefully and answer the questions
In XI standard, teacher was discussing about Spearman's Rank Correlation Coefficient in Statistics. Following points were discussed in the class about the same topic:
Image
When we are finding correlation between two qualitative characteristics, say, beauty and intelligence, we take recourse to using rank correlation coefficient. Rank correlation can also be applied to find the level of agreement (or disagreement) between two judges so far as assessing a qualitative characteristic is concerned. As compared to product moment correlation coefficient, rank correlation is easier to compute, it can also be advocated to get a first hand impression about the correlation between a pair of variables. Spearman's rank correlation coefficient is given by
$r _{ S }=1-\frac{6 \Sigma d^2}{n\left(n^2-1\right)}$
where,
d = difference between ranks of corresponding x and y
n = number of pairs of values (x,y) in the data
When the rank are repeated, the Spearman's rank correlation coefficient formula is given by
$r _{ S }=1-\frac{6\left[\Sigma d^2+\frac{\left(m_1^3-m_1\right)}{12}+\frac{\left(m_2^3-m_2\right)}{12}+\ldots \ldots\right]}{n\left(n^2-1\right)}$
where, $m _1, m_2 \ldots$, are the number of repetitions of ranks and $\frac{m_1^3-m_1}{12} \ldots$ their corresponding correlation factors.
For example: Ranks obtained by 10 students are given below:

34789
73411

(a) Find the value of $\Sigma d$?
(b) Find the value $\Sigma d^2$?
(c) What is the value of $n^2$ in the given data?
OR
What is rank correlation of the given data?

Answer
In XI standard, teacher was discussing about Spearman's Rank Correlation Coefficient in Statistics. Following points were discussed in the class about the same topic:
Image
When we are finding correlation between two qualitative characteristics, say, beauty and intelligence, we take recourse to using rank correlation coefficient. Rank correlation can also be applied to find the level of agreement (or disagreement) between two judges so far as assessing a qualitative characteristic is concerned. As compared to product moment correlation coefficient, rank correlation is easier to compute, it can also be advocated to get a first hand impression about the correlation between a pair of variables. Spearman's rank correlation coefficient is given by
$r _{ S }=1-\frac{6 \Sigma d^2}{n\left(n^2-1\right)}$
where,
d = difference between ranks of corresponding x and y
n = number of pairs of values (x,y) in the data
When the rank are repeated, the Spearman's rank correlation coefficient formula is given by
$r _{ S }=1-\frac{6\left[\Sigma d^2+\frac{\left(m_1^3-m_1\right)}{12}+\frac{\left(m_2^3-m_2\right)}{12}+\ldots \ldots\right]}{n\left(n^2-1\right)}$
where, $m _1, m_2 \ldots$, are the number of repetitions of ranks and $\frac{m_1^3-m_1}{12} \ldots$ their corresponding correlation factors.
For example: Ranks obtained by 10 students are given below:
34789
73411
(i)
Rank of Maths $\left( R _{ x }\right)$Rank of Physics ( $R _{ y }$ )$d = R _{ x }- R _{ y }$
1055
56-1
69-3
12-1
28-6
37-4
431
743
81-2
918
--0
Therefore, $\Sigma d=0$
(ii) $\Sigma d^2=25+1+9+1+36+16+1+9+4+64$
= 166
(ii) $n^2=10^2$
= 100
OR
Rank correlation is given by:
$\begin{array}{l} r _{ S }=1-\frac{6 \Sigma d^2}{n\left(n^2-1\right)} \\ =1-\frac{6 \times 166}{10\left(10^2-1\right)}\end{array}$
$\begin{array}{l}=1-\frac{996}{990} \\ =\frac{990-996}{990} \\ =\frac{-6}{990} \\ =-0.006\end{array}$
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Question 44 Marks
Answer
A market is in the form of a triangle whose vertices are B(-2, 0), C(1, 12). The third vertex A of this trianagle lies on the mid point of the line joining the points (2, 1) and (4, 13).
Image
(i) Midpoint of (2, 1) and (4, 13) is given by (3, 7).
(ii) Slope $=\frac{12-0}{1-(-2)}=\frac{12}{3}=4$
(iii)Equation of BC: $=(y-0)=4(x+2)$
$=y-2=4 x$
OR
Equation of BC is y - 2 - 4x = 0. Putting coordinates of A (3, 7) in this equation, we get: $7-2-12 \neq 0$, therefore we can say that point A will not lie on the line BC.
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Case study (4 Marks) - Applied Maths STD 11 Science Questions - Vidyadip