Question

Run the following R commends.
set.seed(2019) x = rnorm(100,mean=0, sd=2) y = rnorm(100,mean=0, sd=20)
[NOTE: Do NOT use built-in functions which can directly solve the problems.]
(1) Compute the means of values in the vectors x and y, respectively. Which one is more close to zero? Explain the reason.
(2) Compute the sample standard deviations of values in the vector x and y, respectively.
(3) Find the index positions in x and y of the values between -5 and 5, respectively

Answer 1

1. Computing the mean value of vector x and y :

We know that the mean = sum of all elements in the array / length of the array.

  For Example : Mean o values in array a = [1,2,3,4,5] is :

mean = (1+2+3+4+5) / 5 = 3

Let's solve the question using R :

#Given  

set.seed(2019)

x = rnorm(100,mean=0, sd=2)

y = rnorm(100,mean=0, sd=20)

#Computin mean of x

mean_x = sum(x) / length(x)   # here sum function add all element of the vector and length function fetch the length vector

#Computing the mean of y

mean_y = sum(y) / length(y)

#Printing the value of mean_x and mean_y

prinit(mean_x) # -0.146668

print(mean_y)   #-3.404389

--> Here we get the mean of vector x close to zero because of the small standard deviation. A smaller standard deviation indicates that more of the data is clustered about the mean.

----------------------------------------------------------------------------------------------------------------------------------------------------------------------

2. Computing the standard deviation for vector x and y :

#Steps to Compute Standard Deviation :   

1. Calculate the mean.
2. Subtract the mean from each observation.
3. Square each of the resulting observations.
4. Add these squared results together.
5. Divide this total by the number of observation
6. Use the positive square root

#Here we will use the value of variable mean_x and mean_y   

sd_x = sqrt(sum((x-mean_x)^2) / (length(x) - 1))

sd_y =   sqrt(sum((y-mean_y)^2) / (length(y) - 1))

  #printing the sd values of x and y

print(sd_x) #1.810907

print(sd_y) #19.7784

------------------------------------------------------------------------------------------------------------------------------------------------------------------

3. Finding the index for the value between -5 to 5 :

# simple code

  which( x == x[ x>= -5 && x<=5])

# [x>= -5 && x<=5] will fetch the value between -5 to 5 from vector x

# We then compare the original array to the boolean indexed vector

# which function simply print the position of those vector

#Similarly for y

  which( y == y[ y>= -5 && y<=5])

----------------------------------------------------------------------------------------------------------------------------------------------------------------------