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Cover image of IGNOU MST-015 Solved Assignment 2023
Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MST-015 Assignment Question Paper 2023

untitled-document-16-9cbde385-ad0a-4c74-864f-362c1a2d74cc
  1. Attempt the following:
    (a) Write the output of the following statements:
(i) rep ⁡ ( X = C ( T , F , T , F ) rep ⁡ ( X = C ( T , F , T , F ) rep(X=C(T,F,T,F)\operatorname{rep}(\mathrm{X}=\mathrm{C}(\mathrm{T}, \mathrm{F}, \mathrm{T}, \mathrm{F})rep⁡(X=C(T,F,T,F), times = C ( 2 , 1 , 2 , 3 ) ) = C ( 2 , 1 , 2 , 3 ) ) =C(2,1,2,3))=\mathrm{C}(2,1,2,3))=C(2,1,2,3))
(ii) 5 % / % 3 5 % / % 3 5%//%35 \% / \% 35%/%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 5,10,8,7,8,5,8,7,5,8,9,6,8,8,85,10,8,7,8,5,8,7,5,8,9,6,8,8,85,10,8,7,8,5,8,7,5,8,9,6,8,8,8
(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<-0x<−0
repeat {
print ⁡ ( x ∧ 2 ) print ⁡ ( x ∧ 2 ) print(x^^2)\operatorname{print}(x \wedge 2)print⁡(x∧2)
x < − x + 1 x < − x + 1 x < -x+1x<-x+1x<−x+1
if ( x < 5 ) print ⁡ ( x ) } ( x < 5 ) print ⁡ ( x ) } (x < 5)print(x)}(x<5) \operatorname{print}(x)\}(x<5)print⁡(x)}
x < − 0 x < − 0 x < -0\mathrm{x}<-0x<−0
repeat {
print ⁡ ( x ∧ 2 ) print ⁡ ( x ∧ 2 ) print(x^^2)\operatorname{print}(x \wedge 2)print⁡(x∧2)
x < − x + 1 x < − x + 1 x < -x+1\mathrm{x}<-\mathrm{x}+1x<−x+1
if ( x < 5 ) ( x < 5 ) (x < 5)(x<5)(x<5) print ( x ) } ( x ) } (x)}(x)\}(x)}
( 1 × 3 + 2 = 5 ) ( 1 × 3 + 2 = 5 ) (1xx3+2=5)(1 \times 3+2=5)(1×3+2=5)
  1. The following data relates to the number of items produced per shift by two workers for a number of days.
Worker A192224272418
Worker B26374035NANA
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 R R R\mathrm{R}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.
( − 2 4 0 1 9 2 ) − 2 4 0 1 9 2 ([-2,4],[0,1],[9,2])\left(\begin{array}{cc} -2 & 4 \\ 0 & 1 \\ 9 & 2 \end{array}\right)(−240192)
Also, extract the row shown in the rectangular box.
(d) Create the graph of the following function.
f ( x ) = | x | , − 5 ≤ x ≤ 5 f ( x ) = | x | , − 5 ≤ x ≤ 5 f(x)=|x|,-5 <= x <= 5f(x)=|x|,-5 \leq x \leq 5f(x)=|x|,−5≤x≤5
  1. (a) Create following two matrices A A AAA and B B BBB with following elements.
A = ( 1 2 0 1 ) , B = ( − 3 1 2 − 1 ) A = 1 2 0 1 , B = − 3 1 2 − 1 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=(1201),B=(−312−1)
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 ∗ B A 2 + 3 ∗ B A^(2)+3^(**)BA^2+3^* BA2+3∗B
(b) Create a data frame named RData consisting of the following data:
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  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 {:[” 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} x  Y  z  W 0.040.160.53 A0.820.870.84 A0.320.650.99 A0.390.830.42 A0.310.930.78 A0.830.310.41 A0.730.740.88 A0.320.390.50 B0.600.850.68 B0.650.280.86 B0.550.950.77 B0.530.350.32 B0.910.010.91 B0.840.810.37 B
Write R R R\mathrm{R}R commands to:
(i) Compute the group wise means of x , y x , y x,y\mathrm{x}, \mathrm{y}x,y and z z z\mathrm{z}z according to the groups defined by w column using apply family function.
(ii) Sort RData according the y column of it.
\(b=c\:cos\:A+a\:cos\:C\)

MST-015 Sample Solution 2023

untitled-document-16-9cbde385-ad0a-4c74-864f-362c1a2d74cc

Question:-01

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

(a) Write the output of the following statements:

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

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

(ii) 5 % / % 3 5 % / % 3 5%//%35 \% / \% 35%/%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.
Answer:
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.
\(sin\left(\theta -\phi \right)=sin\:\theta \:cos\:\phi -cos\:\theta \:sin\:\phi \)
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\(sin\left(2\theta \right)=2\:sin\:\theta \:cos\:\theta \)

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