Answer to Question #120428 in Statistics and Probability for Lowe kris

Question

2. Suppose that the random variable, X, is a number on the biased die and the p.d.f. of X is as shown
below;
X 1 2 3 4 5 6
P(X=x) 1/6 1/6 1/5 k 1/5 1/6
a) Find;
(i) the value of k.
(ii) E(X)
(iii) E(X2
)
(iv) Var(X)
(v) P(1 X <5)

Answer 1

We have,

X = random variable denoting the number on the biased die

The probability mass function of X is given by,

X 1 2 3 4 5 6

P(X=x) 1/6 1/6 1/5 k 1/5 1/6

a) (i) We know the total probability is always 1.

$\therefore \displaystyle \sum_{x=1}^6P(X=x)=1$

i.e. $\frac{1}{6}+\frac{1}{6}+\frac{1}{5}+k+\frac{1}{5}+\frac{1}{6}=1$

i.e. $k+\frac{9}{10}=1$

i.e. $k=1-\frac{9}{10}=\frac{1}{10}$

Answer: The value of k is $\frac{1}{10}$.

(ii) $E(X)=\displaystyle \sum_{x=1}^6x.P(X=x)= 1.\frac{1}{6}+2.\frac{1}{6}+3.\frac{1}{5}+4.\frac{1}{10}+5.\frac{1}{5}+6.\frac{1}{6}=\frac{7}{2}$

Answer: E(X) = $\frac{7}{2}$.

(iii) $E(X^2)=\displaystyle \sum_{x=1}^6x^2.P(X=x)= 1^2.\frac{1}{6}+2^2.\frac{1}{6}+3^2.\frac{1}{5}+4^2.\frac{1}{10}+5^2.\frac{1}{5}+6^2.\frac{1}{6}=\frac{457}{30}$

Answer: E(X²) = $\frac{457}{30}$.

(iv) $Var(X)=E(X^2)-[E(X)]^2=\frac{457}{30}-(\frac{7}{2})^2=\frac{179}{60}$

Answer: Var(X) = $\frac{179}{60}$.

(v) P(1 < X < 5) [the sign between 1 and X is missing in the question. We assume it is less than (<)]

= P(X = 2) + P(X = 3) + P(X = 4)

= $\frac{1}{6}+\frac{1}{5}+\frac{1}{10}$

= $\frac{7}{15}$

Answer: P(1 < X < 5) = $\frac{7}{15}$.