UNISA
STA2601 Assignment 1 Solutions 2026
Unique: 594831
Opening date: 24 April 2026
Due date: 11 May 2016
, Question 1
Given:
𝑦2 − 1
𝑃(𝑌 = 𝑦) = , 𝑦 = 2,3,4,5
𝑚
1.1 Determine 𝒎
Since probabilities must sum to 1:
∑𝑃(𝑌 = 𝑦) = 1
(22 − 1) + (32 − 1) + (42 − 1) + (52 − 1)
=1
𝑚
Calculate each term:
22 − 1 = 3
32 − 1 = 8
42 − 1 = 15
52 − 1 = 24
Sum:
3 + 8 + 15 + 24 = 50
So:
50
= 1 ⇒ 𝑚 = 50
𝑚
1.2 Probability distribution of 𝒀
𝑦2 − 1
𝑃(𝑌 = 𝑦) =
50
𝑦 𝑃(𝑌 = 𝑦)
2 3/50
3 8/50
STA2601 Assignment 1 Solutions 2026
Unique: 594831
Opening date: 24 April 2026
Due date: 11 May 2016
, Question 1
Given:
𝑦2 − 1
𝑃(𝑌 = 𝑦) = , 𝑦 = 2,3,4,5
𝑚
1.1 Determine 𝒎
Since probabilities must sum to 1:
∑𝑃(𝑌 = 𝑦) = 1
(22 − 1) + (32 − 1) + (42 − 1) + (52 − 1)
=1
𝑚
Calculate each term:
22 − 1 = 3
32 − 1 = 8
42 − 1 = 15
52 − 1 = 24
Sum:
3 + 8 + 15 + 24 = 50
So:
50
= 1 ⇒ 𝑚 = 50
𝑚
1.2 Probability distribution of 𝒀
𝑦2 − 1
𝑃(𝑌 = 𝑦) =
50
𝑦 𝑃(𝑌 = 𝑦)
2 3/50
3 8/50
