A random variable X has the following probability distribution : X 0 1 2 3 4 5 6 7 P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k Determine: k P(X < 3) P( X > 4) - Mathematics and Statistics

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Sum

A random variable X has the following probability distribution:

X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k

Determine:

  1. k
  2. P(X < 3)
  3. P( X > 4)
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Solution 1

i. Since P(x) is a probability distribution of X,

`Sigma_(x = 0)^7 P(x) = 1`

∴ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1

∴ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1

∴ 10k2 + 9k − 1 = 0

∴ 10k2 + 10k − k − 1 = 0

∴ 10k (k + 1) − 1(k + 1) = 0

∴ (k + 1)(10k − 1) = 0

∴ 10k − 1 = 0

∴ 10k − 1 = 0    ........(k ≠ −1)

∴ k =`1/10`

ii. P(X < 3) = P(0) + P(1) + P(2)

= 0 + k + 2k

= 3k

= `3 (1/10)`

= `3/10`

iii. P(0 < X < 3) = + P(1) + P(2)

= k + 2k

= 3k

= `3(1/10)`

= `3/10`.

Solution 2

i. The table gives a probability distribution and therefore `sum_(i = 1)^8 P_i` = 1

∴ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1

∴ 10k2 + 9k – 1 = 0

∴ 10k2 + 10k – k – 1 = 0

∴ 10k(k + 1) – 1(k + 1) = 0

∴ (10k – 1)(k + 1) = 0

∴ k = `1/10` or k = –1

But k cannot be negative

∴ k = `1/10`

ii. P(X < 3)

= P(X = 0 or X = 1 or X = 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

= 3k

= `3/10` 

iii. P(X > 4)

= P(X = 5 or X = 6 or X = 7)

= P(X = 5) + P(X = 6) + P(X = 7)

= k2 + 2k2 + 7k2 + k

= 10k2 + k

= `10(1/10)^2 + 1/10`

= `1/10 + 1/10`

= `1/5`

Concept: Probability Distribution of Discrete Random Variables
  Is there an error in this question or solution?
Chapter 7: Probability Distributions - Exercise 7.1 [Page 232]

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