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We have, n = 20, and σ2 = 5
Now each observation is multiplied by 2.
Suppose X = 2x be the new data.
Since, σ2 = 5
Now, for the new data:
We have, n = 15, and σ2 = 4
Now each observation is increased by 9.
Suppose x = x + 9 be the new data.
Let the other two be x and y
1 + 2 + 6 + x + y = 5 * 4.4 because of the mean
x + y = 13
Variance = [(1 - 4.4)2 + (2 - 4.4)2 + (6 - 4.4)2 + (x - 4.4)2 + (y - 4.4)2]/5
Hence
11.56 + 5.76 + 2.56 + (x - 4.4)2 + (y - 4.4)2 = 41.2
(x - 4.4)2 + (13 - x - 4.4)2 = 21.32
x2 - 8.8x + 19.36 + 73.96 - 17.2x + x2 = 21.32
2x2 - 26x + 72 = 0
x2 - 13x + 36 = 0
(x - 4)(x - 9) = 0
x = 4 or x = 9
If x = 4, y = 9 and
The other two observations are 4 and 9.
If mean and SD of observations areand σ respectively, then mean and SD of observations multiplied by a constant 'k' are
SD = |k|σ
In this question, it is given that k = 3
So New mean = 8 × 3 = 24
New SD = 4 × 3 = 12
Let x and y be the remaining two observations. Then,
Mean = 9
⟹ 60 + x + y + = 72
⟹ x + y = 12 ..... (i)
Variance = 9.25
⟹ 642 + x2 + y2 = 722
⟹ x2 + y2 = 80 ..... (ii)
Now, (x + y)2 + (x - y)2 = 2(x2 + y2)
⟹ 144 + (x - y)2 = 2 × 80
⟹ (x - y)2 = 16
⟹ x - y = ±4
if x - y = 4, then x + y = 12 and x - y = 4 ⟹ x = 8, y = 4
if x - y = -4, then x + y = 12 and x - y = -4 ⟹ x = 4, y = 8
Hence, the remaining two observations are 4 and 8.
We have,
n = 200, = 40, σ = 15
Corrected Σxi = Incorrect Σxi - (Sum of incorrect values) + (Sum of correct values)
= 8000 - 34 - 53 + 43 + 35 = 7991
= 365000 - (34)2 - 532 + (43)2 + 352 = 364109
891
n = 100,= 40, σ = 5.1
Corrected Σxi = incorrect Σxi - (Sum of incorrect values) + (Sum of correct values)
= 4000 - 50 + 40 = 3990
= 162601 - (50)2 + (40)2 = 161701
n = 20,= 10 and σ = 2
⟹ Incorrected Σxi = 200
and,
(i) When 8 is omitted from the data:
If 8 is omitted from the data, then 19 observations are left.
Now, Incorrected Σxi = 200
⟹ Corrected Σxi + 8 = 200
⟹ Corrected Σxi = 192
(ii) When the incorrect observation 8 is replaced by 12:
we have, Incorrected Σxi = 200
∴ Corrected Σxi = 200 - 8 + 12 = 204
⟹ Incorrect Σxi = 2000
When the incorrect observations 21, 21, 18 are omitted from the data:
n = 97
Now, Incorrect Σxi = 2000
⟹ Corrected Σxi = 2000 - 21 - 21 - 18 = 1940
Dividing both the sides by n we get,
Taking square root on both the sides
Chapter 32: Statistics – Exercise 32.6...
Chapter 32: Statistics – Exercise 32.7...
Chapter 32: Statistics – Exercise 32.5...
Chapter 32: Statistics – Exercise 32.3...
Chapter 32: Statistics – Exercise 32.2...
Chapter 32: Statistics – Exercise 32.1...