IGNOU MST-015 Solved Assignment 2023 | MSCAST

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

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IGNOU MST-015 Assignment Question Paper 2023

1. Attempt the following:
(a) Write the output of the following statements:
(i) $\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$$\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$rep(X=C(T,F,T,F)\operatorname{rep}(\mathrm{X}=\mathrm{C}(\mathrm{T}, \mathrm{F}, \mathrm{T}, \mathrm{F})$\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$, times $=\mathrm{C}\left(2,1,2,3\right)\right)$$=\mathrm{C}\left(2,1,2,3\right)\right)$=C(2,1,2,3))=\mathrm{C}(2,1,2,3))$=\mathrm{C}\left(2,1,2,3\right)\right)$
(ii) $5\mathrm{%}/\mathrm{%}3$$5\mathrm{%}/\mathrm{%}3$5%//%35 \% / \% 3$5\mathrm{%}/\mathrm{%}3$; diag (3)
(b) Differentiate between the use of the sep and col1 apse arguments of the paste() function.
(c) Write R commands to create a bar plot of the following data by using arguments of the used function for filling up the bars and to give labels to the axis:
$5,10,8,7,8,5,8,7,5,8,9,6,8,8,8$$5,10,8,7,8,5,8,7,5,8,9,6,8,8,8$5,10,8,7,8,5,8,7,5,8,9,6,8,8,85,10,8,7,8,5,8,7,5,8,9,6,8,8,8$5,10,8,7,8,5,8,7,5,8,9,6,8,8,8$
(d) Check whether the given loop is finite or infinite. If infinite, do the necessary changes in the written loop to make it finite.
$x<-0$$x<-0$x < -0x<-0$x<-0$
repeat {
$\mathrm{print}\left(x\wedge 2\right)$$\mathrm{print}\left(x\wedge 2\right)$print(x^^2)\operatorname{print}(x \wedge 2)$\mathrm{print}\left(x\wedge 2\right)$
$x<-x+1$$x<-x+1$x < -x+1x<-x+1$x<-x+1$
if $\left(x<5\right)\mathrm{print}\left(x\right)\right\}$$\left(x<5\right)\mathrm{print}\left(x\right)\right\}$(x < 5)print(x)}(x<5) \operatorname{print}(x)\}$\left(x<5\right)\mathrm{print}\left(x\right)\right\}$
$\mathrm{x}<-0$$\mathrm{x}<-0$x < -0\mathrm{x}<-0$\mathrm{x}<-0$
repeat {
$\mathrm{print}\left(x\wedge 2\right)$$\mathrm{print}\left(x\wedge 2\right)$print(x^^2)\operatorname{print}(x \wedge 2)$\mathrm{print}\left(x\wedge 2\right)$
$\mathrm{x}<-\mathrm{x}+1$$\mathrm{x}<-\mathrm{x}+1$x < -x+1\mathrm{x}<-\mathrm{x}+1$\mathrm{x}<-\mathrm{x}+1$
if $\left(x<5\right)$$\left(x<5\right)$(x < 5)(x<5)$\left(x<5\right)$ print $\left(x\right)\right\}$$\left(x\right)\right\}$(x)}(x)\}$\left(x\right)\right\}$
$\left(1×3+2=5\right)$$\left(1×3+2=5\right)$(1xx3+2=5)(1 \times 3+2=5)$\left(1×3+2=5\right)$
1. The following data relates to the number of items produced per shift by two workers for a number of days.
 Worker A 19 22 24 27 24 18 Worker B 26 37 40 35 NA NA
Worker A 19 22 24 27 24 18 Worker B 26 37 40 35 NA NA| Worker A | 19 | 22 | 24 | 27 | 24 | 18 | | :— | :— | :— | :— | :— | :— | :— | | Worker B | 26 | 37 | 40 | 35 | NA | NA |
(a) Write R command to create a list named LT with worker’s data. Also, after creating the list, do the following tasks:
(i) Use a suitable loop function to compute the mean of number of items produced by each worker in a single line command.
(ii) Extract the worker A data from it by using two different approaches.
(b) Write $\mathrm{R}$$\mathrm{R}$R\mathrm{R}$\mathrm{R}$ command to create a data frame named DF with worker’s data and do the following tasks:
(i) Use suitable function to remove NA from the data and then create a scatter plot.
(ii) Write the known data obtained in step (i) to a .txt file named “WORK”.
1. Write R commands to:
(a) Create a function to compute ranks (in case of tied ranks) of the given data.
(b) Create a date object named Ddata consisting of the following dates.
26Jan2023, 15Aug2023, 02Oct2023, 05Sep2023
(c) Create an array of two dimension with following elements.
$\left(\begin{array}{cc}-2& 4\\ 0& 1\\ 9& 2\end{array}\right)$$\left(\begin{array}{cc}-2& 4\\ 0& 1\\ 9& 2\end{array}\right)$([-2,4],[0,1],[9,2])\left(\begin{array}{cc} -2 & 4 \\ 0 & 1 \\ 9 & 2 \end{array}\right)$\left(\begin{array}{cc}-2& 4\\ 0& 1\\ 9& 2\end{array}\right)$
Also, extract the row shown in the rectangular box.
(d) Create the graph of the following function.
$f\left(x\right)=|x|,-5\le x\le 5$$f\left(x\right)=|x|,-5\le x\le 5$f(x)=|x|,-5 <= x <= 5f(x)=|x|,-5 \leq x \leq 5$f\left(x\right)=|x|,-5\le x\le 5$
1. (a) Create following two matrices $A$$A$AA$A$ and $B$$B$BB$B$ with following elements.
$A=\left(\begin{array}{ll}1& 2\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}-3& 1\\ 2& -1\end{array}\right)$$A=\left(\begin{array}{ll}1& 2\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}-3& 1\\ 2& -1\end{array}\right)$A=([1,2],[0,1]),B=([-3,1],[2,-1])A=\left(\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right), B=\left(\begin{array}{cc} -3 & 1 \\ 2 & -1 \end{array}\right)$A=\left(\begin{array}{ll}1& 2\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}-3& 1\\ 2& -1\end{array}\right)$
Write R commands to do the following tasks:
(i) Multiply the two matrices.
(ii) Combine the two matrices row-wise.
(iii) Create a function that computes the following expression:
${A}^{2}+{3}^{\ast }B$${A}^{2}+{3}^{\ast }B$A^(2)+3^(**)BA^2+3^* B${A}^{2}+{3}^{\ast }B$
(b) Create a data frame named RData consisting of the following data:
$\begin{array}{rrrr}\text{x}& \text{Y}& \text{z}& \text{W}\\ 0.04& 0.16& 0.53& \text{}\mathrm{A}\\ 0.82& 0.87& 0.84& \text{}\mathrm{A}\\ 0.32& 0.65& 0.99& \text{}\mathrm{A}\\ 0.39& 0.83& 0.42& \text{}\mathrm{A}\\ 0.31& 0.93& 0.78& \text{}\mathrm{A}\\ 0.83& 0.31& 0.41& \text{}\mathrm{A}\\ 0.73& 0.74& 0.88& \text{}\mathrm{A}\\ 0.32& 0.39& 0.50& \text{}\mathrm{B}\\ 0.60& 0.85& 0.68& \text{}\mathrm{B}\\ 0.65& 0.28& 0.86& \text{}\mathrm{B}\\ 0.55& 0.95& 0.77& \text{}\mathrm{B}\\ 0.53& 0.35& 0.32& \text{}\mathrm{B}\\ 0.91& 0.01& 0.91& \text{}\mathrm{B}\\ 0.84& 0.81& 0.37& \text{}\mathrm{B}\end{array}${:[” x “,” Y “,” z “,” W “],[0.04,0.16,0.53,A],[0.82,0.87,0.84,A],[0.32,0.65,0.99,A],[0.39,0.83,0.42,A],[0.31,0.93,0.78,A],[0.83,0.31,0.41,A],[0.73,0.74,0.88,A],[0.32,0.39,0.50,B],[0.60,0.85,0.68,B],[0.65,0.28,0.86,B],[0.55,0.95,0.77,B],[0.53,0.35,0.32,B],[0.91,0.01,0.91,B],[0.84,0.81,0.37,B]:}\begin{array}{rrrr} \text { x } & \text { Y } & \text { z } & \text { W } \\ 0.04 & 0.16 & 0.53 & \mathrm{~A} \\ 0.82 & 0.87 & 0.84 & \mathrm{~A} \\ 0.32 & 0.65 & 0.99 & \mathrm{~A} \\ 0.39 & 0.83 & 0.42 & \mathrm{~A} \\ 0.31 & 0.93 & 0.78 & \mathrm{~A} \\ 0.83 & 0.31 & 0.41 & \mathrm{~A} \\ 0.73 & 0.74 & 0.88 & \mathrm{~A} \\ 0.32 & 0.39 & 0.50 & \mathrm{~B} \\ 0.60 & 0.85 & 0.68 & \mathrm{~B} \\ 0.65 & 0.28 & 0.86 & \mathrm{~B} \\ 0.55 & 0.95 & 0.77 & \mathrm{~B} \\ 0.53 & 0.35 & 0.32 & \mathrm{~B} \\ 0.91 & 0.01 & 0.91 & \mathrm{~B} \\ 0.84 & 0.81 & 0.37 & \mathrm{~B} \end{array}
Write $\mathrm{R}$$\mathrm{R}$R\mathrm{R}$\mathrm{R}$ commands to:
(i) Compute the group wise means of $\mathrm{x},\mathrm{y}$$\mathrm{x},\mathrm{y}$x,y\mathrm{x}, \mathrm{y}$\mathrm{x},\mathrm{y}$ and $\mathrm{z}$$\mathrm{z}$z\mathrm{z}$\mathrm{z}$ according to the groups defined by w column using apply family function.
(ii) Sort RData according the y column of it.
$$c=a\:cos\:B+b\:cos\:A$$

MST-015 Sample Solution 2023

Question:-01

1. Attempt the following:
(a) Write the output of the following statements:
(i) $\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$$\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$rep(X=C(T,F,T,F)\operatorname{rep}(\mathrm{X}=\mathrm{C}(\mathrm{T}, \mathrm{F}, \mathrm{T}, \mathrm{F})$\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$, times $=\mathrm{C}\left(2,1,2,3\right)\right)$$=\mathrm{C}\left(2,1,2,3\right)\right)$=C(2,1,2,3))=\mathrm{C}(2,1,2,3))$=\mathrm{C}\left(2,1,2,3\right)\right)$
(ii) $5\mathrm{%}/\mathrm{%}3$$5\mathrm{%}/\mathrm{%}3$5%//%35 \% / \% 3$5\mathrm{%}/\mathrm{%}3$; diag (3)

(a) Write the output of the following statements:

(i) $\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$$\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$rep(X=C(T,F,T,F)\operatorname{rep}(\mathrm{X}=\mathrm{C}(\mathrm{T}, \mathrm{F}, \mathrm{T}, \mathrm{F})$\mathrm{rep}\left(\mathrm{X}=\mathrm{C}\left(\mathrm{T},\mathrm{F},\mathrm{T},\mathrm{F}\right)$, times $=\mathrm{C}\left(2,1,2,3\right)\right)$$=\mathrm{C}\left(2,1,2,3\right)\right)$=C(2,1,2,3))=\mathrm{C}(2,1,2,3))$=\mathrm{C}\left(2,1,2,3\right)\right)$

In R, this statement would replicate the elements of the vector $X=\text{c}\left(T,F,T,F\right)$$X=\text{c}\left(T,F,T,F\right)$X=”c”(T,F,T,F)X = \text{c}(T, F, T, F)$X=\text{c}\left(T,F,T,F\right)$ according to the times argument $\text{c}\left(2,1,2,3\right)$$\text{c}\left(2,1,2,3\right)$“c”(2,1,2,3)\text{c}(2, 1, 2, 3)$\text{c}\left(2,1,2,3\right)$.
The output would be a logical vector: $\text{c}\left(T,T,F,T,T,F,F,F\right)$$\text{c}\left(T,T,F,T,T,F,F,F\right)$“c”(T,T,F,T,T,F,F,F)\text{c}(T, T, F, T, T, F, F, F)$\text{c}\left(T,T,F,T,T,F,F,F\right)$

(ii) $5\mathrm{%}/\mathrm{%}3$$5\mathrm{%}/\mathrm{%}3$5%//%35 \% / \% 3$5\mathrm{%}/\mathrm{%}3$; diag (3)

In R, 5 %% 3 would calculate the remainder of 5 divided by 3, which is 2.
The diag(3) would create a 3×3 identity matrix.
The output would be:
[1] 2
[,1] [,2] [,3]
[1,]    1    0    0
[2,]    0    1    0
[3,]    0    0    1


(b) Differentiate between the use of the sep and collapse arguments of the paste() function.
In R, the paste() function is used for concatenating strings. The sep and collapse arguments serve different purposes within this function:

sep Argument:

The sep argument specifies the string that separates the elements to be concatenated. It is used to separate multiple pieces of text that you are joining together.
For example:
paste("Hello", "World", sep = " ")

Output: "Hello World"
Here, the sep argument puts a space between “Hello” and “World”.

collapse Argument:

The collapse argument is used when you have a vector of strings that you want to collapse into a single string. It specifies the separator between elements within the resulting single string.
For example:
paste(c("Hello", "World"), collapse = " ")

Output: "Hello World"
Here, the collapse argument collapses the vector c("Hello", "World") into a single string, separating the elements by a space.

Key Differences:

1. sep is used to separate the elements that are being concatenated in a single call to paste().
2. collapse is used to collapse a vector of strings into a single string.
3. sep works between the elements in each function call, whereas collapse works on the entire vector to produce a single string.
Here’s an example that uses both sep and collapse:
paste(c("Hello", "World"), c("How", "are", "you"), sep = "-", collapse = " ")

Output: "Hello-How World-are Hello-you"
In this example, sep = "-" puts a hyphen between elements being concatenated (“Hello” with “How”, “World” with “are”, and so on). Then collapse = " " collapses the resulting vector into a single string, separating each element by a space.
$$b^2=c^2+a^2-2ac\:Cos\left(B\right)$$

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