The discrete random variable X has probability distribution | x | -1 | 0 | 1 | 2 | |------|----|---|---|---| | P(X = x) | a | b | b | c | The cumulative distribution function of X is given by | x | -1 | 0 | 1 | 2 | |------|----|---|---|---| | F(x) | 1/3 | d | 5/6 | e | (a) Find the values of a, b, c, d and e - Edexcel - A-Level Maths Statistics - Question 4 - 2017 - Paper 1

To determine the values of a, b, c, d, and e, we can use the properties of probability distribution and cumulative distribution functions:

1. Sum of Probabilities: The total probability must sum to 1:

   $a + b + b + c = 1$

   This can be simplified to:

   $a + 2b + c = 1$

2. Cumulative Distribution Function: The cumulative distribution function values must satisfy:

   • For x = -1: ( F(-1) = a = 1/3 )
   • For x = 0: ( F(0) = a + b = d )
   • For x = 1: ( F(1) = a + 2b = 5/6 )
   • For x = 2: ( F(2) = a + 2b + c = e )

From the first equation, we already found: ( a = 1/3 )

Substituting ( a ) into the second and third equations gives:

3. Deriving b and c:

   • From ( F(0) ):
   • ( 1/3 + b = d ) => Therefore, ( d = 1/3 + b )
   • From ( F(1) ):
   • ( 1/3 + 2b = 5/6 ) => ( 2b = 5/6 - 1/3 )

   Solving: ( 2b = 5/6 - 2/6 ) => ( 2b = 3/6 )

   Thus, ( b = 3/12 = 1/4 )

Substituting ( b ) back in:

• ( d = 1/3 + 1/4 = 4/12 + 3/12 = 7/12 )

4. Solving for c and e:

   We now know ( a ), ( b ), and ( d ).

   • Using ( a + 2b + c = 1 ): ( 1/3 + 2(1/4) + c = 1 ) which simplifies to ( 1/3 + 1/2 + c = 1 )
   • Thus, solving for c gives:
   • ( c = 1 - (1/3 + 1/2) = 1 - (2/6 + 3/6) = 1 - 5/6 = 1/6 )

   Finally, we can find ( e ):

   • ( e = a + 2b + c = 1/3 + 1/2 + 1/6 )
   • To compute this, convert to a common denominator:
   • ( e = 4/12 + 6/12 + 2/12 = 12/12 = 1 )

Therefore, the values are:

• ( a = rac{1}{3} )
• ( b = rac{1}{4} )
• ( c = rac{1}{6} )
• ( d = rac{7}{12} )
• ( e = 1 )