# IGNOU MST-012 Solved Assignment 2023 | MSCAST

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MST-012 Assignment Question Paper 2023

1(a) Suppose two friends Anjali and Prabhat trying to meet for a date to have lunch say between $2\mathrm{p}\mathrm{m}$$2\mathrm{p}\mathrm{m}$2pm2 \mathrm{pm}$2\mathrm{p}\mathrm{m}$ to $3\mathrm{p}\mathrm{m}$$3\mathrm{p}\mathrm{m}$3pm3 \mathrm{pm}$3\mathrm{p}\mathrm{m}$. Suppose they follow the following rules for this meeting:
• Each of them will reach either on time or 10 minutes late or 20 minutes late or $\mathbf{3}\mathbf{0}$$\mathbf{3}\mathbf{0}$30\mathbf{3 0}$\mathbf{3}\mathbf{0}$ minutes late or 40 minutes late or 50 minutes late or 1 hour late. All these arrival times are equally likely for both of them.
• Whoever of them reaches first will wait for the other to meet only for 10 minutes. If within 10 minutes the other does not reach, he/she leaves the place and they will not meet.
Find the probability of their meeting.
(b) In the study learning material (SLM), you have seen many situations where Poisson distribution is suitable and discussed some examples of such situations. Create your own example for a situation other than which are discussed in SLM. If you denote your created random variable by $X$$X$XX$X$ then find the probability that $X$$X$XX$X$ is less than 2 .
2(a) In an election there are two candidates. Being a statistician, you are interested in predicting the result of the election. So, you plan to conduct a survey. Using the learning skill of this course answer the following question. How many people should be surveyed to be at least $90\mathrm{%}$$90\mathrm{%}$90%90 \%$90\mathrm{%}$ sure that the estimate is within 0.03 of the true value?
(b) If ${\mathrm{X}}_{1},{\mathrm{X}}_{2},{\mathrm{X}}_{3},{\mathrm{X}}_{4},\dots$${\mathrm{X}}_{1},{\mathrm{X}}_{2},{\mathrm{X}}_{3},{\mathrm{X}}_{4},\dots$X_(1),X_(2),X_(3),X_(4),dots\mathrm{X}_1, \mathrm{X}_2, \mathrm{X}_3, \mathrm{X}_4, \ldots${\mathrm{X}}_{1},{\mathrm{X}}_{2},{\mathrm{X}}_{3},{\mathrm{X}}_{4},\dots$ be a sequence of continuous random variables such that ${f}_{{x}_{n}}\left(x\right)=\frac{n}{2}{e}^{-n|x|}$${f}_{{x}_{n}}\left(x\right)=\frac{n}{2}{e}^{-n|x|}$f_(x_(n))(x)=(n)/(2)e^(-n|x|)f_{x_n}(x)=\frac{n}{2} e^{-n|x|}${f}_{{x}_{n}}\left(x\right)=\frac{n}{2}{e}^{-n|x|}$ then prove that ${X}_{n}\stackrel{p}{\to }0$${X}_{n}\stackrel{p}{\to }0$X_(n)rarr^(p)0X_n \stackrel{p}{\rightarrow} 0${X}_{n}\stackrel{p}{\to }0$.
1. Explain the procedure of assigning probability in a continuous world of probability theory.
4(a) If $X\sim \mathrm{Gamma}\left(\theta ,\alpha \right)$$X\sim \mathrm{Gamma}\left(\theta ,\alpha \right)$X∼Gamma(theta,alpha)X \sim \operatorname{Gamma}(\theta, \alpha)$X\sim \mathrm{Gamma}\left(\theta ,\alpha \right)$ and $Y\sim \mathrm{Gamma}\left(\theta ,\beta \right)$$Y\sim \mathrm{Gamma}\left(\theta ,\beta \right)$Y∼Gamma(theta,beta)Y \sim \operatorname{Gamma}(\theta, \beta)$Y\sim \mathrm{Gamma}\left(\theta ,\beta \right)$ be two independent gamma distributions and $U=\frac{X}{X+Y}$$U=\frac{X}{X+Y}$U=(X)/(X+Y)U=\frac{X}{X+Y}$U=\frac{X}{X+Y}$ and $V=X+Y$$V=X+Y$V=X+YV=X+Y$V=X+Y$ then find the distribution of $U$$U$UU$U$.
(b) A hospital specialising in heart surgery. In 2022 total of 2000 patients were admitted for treatment. The average payment made by a patient was Rs $1,50,000$$1,50,000$1,50,0001,50,000$1,50,000$ with a standard deviation of Rs 25000 . Under the assumption that payments follow a normal distribution, answer the following questions.
(i) The number of patients who paid between Rs $1,40,000$$1,40,000$1,40,0001,40,000$1,40,000$ and Rs $1,70,000$$1,70,000$1,70,0001,70,000$1,70,000$.
(ii) The probability that a patient bill exceeds Rs $1,00,000$$1,00,000$1,00,0001,00,000$1,00,000$.
(iii) Maximum amount paid by the lowest paying one-third of patients.
$$\operatorname{cosec}^2 \theta=1+\cot ^2 \theta$$

## MEV-017 Sample Solution 2023

### Question:-01

1(a) Suppose two friends Anjali and Prabhat trying to meet for a date to have lunch say between $2\mathrm{p}\mathrm{m}$$2\mathrm{p}\mathrm{m}$2pm2 \mathrm{pm}$2\mathrm{p}\mathrm{m}$ to $3\mathrm{p}\mathrm{m}$$3\mathrm{p}\mathrm{m}$3pm3 \mathrm{pm}$3\mathrm{p}\mathrm{m}$. Suppose they follow the following rules for this meeting:
• Each of them will reach either on time or 10 minutes late or 20 minutes late or $\mathbf{3}\mathbf{0}$$\mathbf{3}\mathbf{0}$30\mathbf{3 0}$\mathbf{3}\mathbf{0}$ minutes late or 40 minutes late or 50 minutes late or 1 hour late. All these arrival times are equally likely for both of them.
• Whoever of them reaches first will wait for the other to meet only for 10 minutes. If within 10 minutes the other does not reach, he/she leaves the place and they will not meet.
Find the probability of their meeting.

### Introduction

Anjali and Prabhat are trying to meet for lunch between 2 pm and 3 pm. They can arrive at seven different times: on time, 10, 20, 30, 40, 50 minutes late, or 1 hour late. Each of these arrival times is equally likely. The first to arrive will wait for 10 minutes for the other. We need to find the probability that they will meet.

### Work/Calculations

#### Step 1: Define the Sample Space

The sample space $S$$S$SS$S$ consists of all possible arrival times for Anjali and Prabhat. Since each of them has 7 choices, the total number of outcomes is $7×7$$7×7$7xx77 \times 7$7×7$.
$|S|=7×7=49$$|S|=7×7=49$|S|=7xx7=49|S| = 7 \times 7 = 49$|S|=7×7=49$

#### Step 2: Identify Favorable Outcomes

A favorable outcome is one where they meet. This can happen if:
1. Both arrive at the same time: There are 7 such cases (0, 10, 20, 30, 40, 50, 60 minutes late).
2. One arrives and the other arrives within the next 10 minutes: For example, if Anjali arrives 10 minutes late, Prabhat can arrive either on time or 20 minutes late for them to meet.
To count these, we consider the time difference between their arrivals. The time difference can be 0, 10, 20, 30, 40, 50, or 60 minutes. They will meet if the time difference is 0 or 10 minutes.
• For a time difference of 0 minutes, there are 7 cases.
• For a time difference of 10 minutes, there are 6 cases for Anjali arriving first and 6 for Prabhat arriving first, making it 12 cases.
So, the total number of favorable outcomes $|E|$$|E|$|E||E|$|E|$ is $7+12=19$$7+12=19$7+12=197 + 12 = 19$7+12=19$.

#### Step 3: Calculate the Probability

The probability $P$$P$PP$P$ of them meeting is given by:
$P=\frac{|E|}{|S|}$$P=\frac{|E|}{|S|}$P=(|E|)/(|S|)P = \frac{|E|}{|S|}$P=\frac{|E|}{|S|}$
After calculating, we get the value of $P$$P$PP$P$.
After calculating, we find that the probability of Anjali and Prabhat meeting is $\frac{19}{49}$$\frac{19}{49}$(19)/(49)\frac{19}{49}$\frac{19}{49}$.

### Conclusion

The probability that Anjali and Prabhat will meet for lunch, given the conditions, is $\frac{19}{49}$$\frac{19}{49}$(19)/(49)\frac{19}{49}$\frac{19}{49}$. This takes into account all the possible times they could arrive and the 10-minute waiting time.
$$cos\:2\theta =2\:cos^2\theta -1$$

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$$b^2=c^2+a^2-2ac\:Cos\left(B\right)$$

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