Answered: "A die is loaded in a such a..."

hroncits8y

Answered question

2021-11-28

A die is loaded in a such a way that probabilities of getting 1, 2, 3, 4, 5, 6 are 1/2, 1/6, 1/12, 1/12, 1/12 and 1/12.
a. Construct the probability distribution for the data.
b. Find the mean and the variance of the probability distribution.

Answer & Explanation

hotcoal3z

Beginner2021-11-29Added 16 answers

Step 1
Provided information is,
A die is loaded in a such a way that probabilities of getting 1, 2, 3, 4, 5, 6 are 1/2, 1/6, 1/12, 1/12, 1/12 and 1/12
1. The probability distribution of the data is:

\begin{array}{cc} x & 1 & 2 & 3 & 4 & 5 & 6 \\ P r o b a b i l i t y & 1 / 2 & 1 / 6 & 1 / 12 & 1 / 12 & 1 / 12 & 1 / 12 \end{array}

P (X = x) = \frac{1}{2} f or x = 1

= \frac{1}{6} f or x = 2

= \frac{1}{12} f or x = 3, 4, 5, 6

Step 2
The mean and variance of the data can be obtained as:

E (x) = \sum_{i = 1}^{n} x . P (X = x)

= (1 \times \frac{1}{2}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{12}) + (4 \times \frac{1}{12}) + (5 \times \frac{1}{12}) + (6 \times \frac{1}{12})

= 2.3333

V a r (x) = E (x^{2}) - {[E (x)]}^{2}

= (1^{2} \times \frac{1}{2}) + (2^{2} \times \frac{1}{6}) + (3^{2} \times \frac{1}{12}) + (4^{2} \times \frac{1}{12}) + (5^{2} \times \frac{1}{12}) + (6^{2} \times \frac{1}{12})

- {[2.3333]}^{2}

= 8.3333 - {[2.3333]}^{2}

= 2.8890

Mean is 2.333 and variance is 2.8890.

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