Test: Statistics- 1

Arranging the given data in ascending order,
we have 34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Here, Median M = (46+48)/2
=47
(∵ n = 10, median is the mean of 5th and 6th items)
∴ Mean deviation = ∑|xi−M|/n
=∑|xi−47|/10
= (13+9+5+3+1+1+7+8+16+23)/10
=8.6

μ = (22 + 26 + 28 + 20 + 24 + 30)/6
= 150/6 
= 25
x(i) = (xi - μ)²
x(22) = (22-25)² = 9
x(26) = (26-25)² = 1
x(28) = (28-25)² = 9
x(20) = (20-25)² = 25
x(24) = (24-25)² = 1
x(30) = (30-25)² = 25
​
(xi - μ)²  = 70
Standard deviation : [(xi - μ)²]/N
= (70/6)^½
= 3.42

ρ = (b(xy) * b(yx))
But sign of ρρ is same as sign of b(xy), b(yx)
Therefore, ρ = 2/7

Given lines are x+4y=3   (i) 
& 3x+y=15        (ii)
We first check which of the line (i) & (ii) is line of regression y
on x and x on y. Let line x+4y=3 be the line of regression x on y, then
other be the line of regression y on x.
∴ From (i) i.e. line of regression x on y we have x=−4y+3
∴ b(xy)=regression coefficient x on y=−4 and 3x+y=15
∴ y=−3x+15
∴ b(yx)=regression coefficient y on x=-3
Now count r²=b(yx)b(xy)=(−4)(−3)=12
⇒ ∣r∣=2(3)^1/2
​which is not possible as 0≤r²≤1. So our assumption is wrong
and line x+4y=3 is line of regression y on x & 3x+y=15 is line of
regression x on y.
∴ 3x+y=15
⇒ x=(15−y)/3
⇒ x=(15−3)/3   (By putting y=3)
​= 4

Let the other nos. be a and b
then (x+y+1+2+6)/5 = 4.4
x + y = 13 ---------------------(1)
Variance = 8.24


41.2 = 19.88 + (x2 + 19.36 – 8.8x) + (y2 + 19.36 – 8.8y) 
21.32 = x2 + y2 + 38.72 – 8.8(x + y) 
x2 + y2 + 38.72 – 8.8(13) – 21.32 = 0 
(using equation (1)) 
x2 + y2 – 97 = 0 …(2) 
Squaring equation (1) both the sides, 
we get (x + y)2 = (13)^2 
x2 + y2 + 2xy = 169 
97 + 2xy = 169 
(using equation (2)) 
xy = 36 or x = 36/y (1)
⇒ 36/y + y = 13 
y2 + 36 = 13y 
y2 – 13y + 36 = 0 
(y – 4)(y – 9) = 0 
Either (y – 4) = 0 or (y – 9) = 0 
⇒ y = 4 or y = 9 
For y = 4 x = 36/y 
= 36/4 = 12 
For y = 9 
x = 36/9 
x = 4 
Thus, remaining two observations are 4 and 9.

The correlation between x and y is positive correlation because it is a correlation where if one variable increases, the other also increases, and if one variable decreases, the other also decreases.

The relationship between the mean, quartile and the standard deviation are as follows:
Mean Deviation is the mean of all the absolute deviations of a set of data.
Quartile deviation is the difference between “first and third quartiles” in any distribution.
Standard deviation measures the “dispersion of the data set” that is relative to its mean.
Mean Deviation = 4/5 × Quartile deviation
Standard Deviation = 3/2 × Quartile deviation

Correlation coefficient = cov (x,y)/ (std deviation (x) ×std deviation (y))
Correlation coefficient  = 0.28
cov (x,y) = 7.6
variance of x is 9.  
=> std deviation (x) = √variance  of X = √9 = 3
=>  0.28  = 7.6 / ( 3 * std deviation (y))
=> std deviation (y) = 7.6 / ( 3 * 0.28)
=> std deviation (y) = 9.05
standard deviation of Y series = 9.05

Median = [mode + 2(mean)]/3
= [6μ+2(9μ)]/3
= 24μ/3
= 8μ