Answer to Question #246735 in Statistics and Probability for strawberry

Question

A discrete random variable 𝑌 has a cumulative probability distribution as the table shown below.

𝑌 =𝑦 0 1 2 3

𝑃(𝑌 ≤ 𝑦) 5𝑘 10𝑘 11𝑘 12𝑘

(a) Determine the value of 𝑘.

(b) Calculate the expected value of 𝑌.

(c) Calculate the variance and standard deviation of 𝑌.

Answer 1

We can calculate the probability as follows:

y    0    1    2   3

y²   0    1    4   9

P   5k  5k  k   k

(a)   Total probability is always 1.

5k + 5k + k + k = 1

k = 1/12

(b)  Expected value = 0 × 5k + 1 × 5k + 2 × k + 3 × k

                        = 10k = 10/12 = 5/6

(c)   Variance = (0 × 5k + 1 × 5k + 4 × k + 9 × k) – (5/6)²

              = 18k – 25/36 = 18/12 – 25/36 = 29/36

Standard deviation = (Variance)^1/2 = (29/36)^1/2 = 0.8975