Probability and Statistics

11 )

If x and y are related as 4x – 3y = 10 and the mean deviation of x is 10 then the mean deviation of y is _______

12.2

13.3

View Solution

Answer :

4x – 3y = 10

∴ 3y = 4x – 10 i.e. y = (4/3)x – (10/3)

∴ MD of y = |(4/3)| MD of x ....... for y = a + bx, MD(y) = |b|MD(x)

∴ MD of y = (4/3) × 10 = (40/3)

∴ MD of y = 13.3.

12 )

For a data ¹⁰∑_i=1 (xi – 15) = 0 and ¹⁰∑_i=1 (xi – 15)² = 160. Then coefficient of variation = ________

24.6

27.5

25.6

26.6

View Solution

Answer :

¹⁰∑_i=1 (xi – 15) = 0

¹⁰∑_i=1 (xi – 15)² = 160

∴ x = 15

n = 10

S.D. = √[{∑(xi – x)²} / n]

= √[{∑(xi – 15)²} / 10]

= √[(160) / (10)]

= 4

∴ S.D. = 4

also x = 15

coefficient of variation = [(S.D.) / x] × 100

= (4/15) × 100

= 26.66%.

13 )

The range of set of 15 observations is 0 then its variance is________

8.25

âˆš15

2.85

View Solution

Answer :

range of observation = 0

∴ R = 0

∴ S = Standard deviation = 0

∴ variance = S² = 0.

14 )

Observations for variable x are 2, 5, 14 and the observations for variable y are 7, 5, 9 then which of the following is true?

Select Correct option from following options

(a) (C.V.)_x > (C.V.)_y (b) (C.V.)_x < (C.V.)_y

View Solution

Answer :

S = √[{(∑xi²) / n} – (x)²] and C.V. = (S / x) × 100

For variable x, x = (21 / 3) = 7 and ∑xi² = 225

∴ S = √[(225 / 3) – 7²] = √26

∴ (C.V.)_x = [√(26) / 7]

Now For y, y = 7 and ∑yi² = 155

∴ S = √[(155 / 3) – 49] = √(8/3)

∴ (C.V.)_y = √(8/3) × (1/7) = [√(8/3) / 7]

∴ (C.V.)_x > (C.V.)_y.

15 )

If n = 100, x = 3 and S² = 11 then [(∑ xi²) / (∑ xi)] is ______

7.55

6.66

5.75

View Solution

Answer :

∑xi = n ∙ x = 100 × 3 = 300

∑(xi)² = n(S² + x²) = 100(11 + 9) = 2000

∴ [(∑xi²) / (∑xi)] = [(2000) / (300)] = (20/3) = 6.66.

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