(a)
By the property of Bernoulli Distribution:
Variance of Bernoulli Distribution is given by:
V(X) = p(1-p)
Given:
V(X) = 0.24
Thus, we get:
p (1 - p) = 0.24
i.e.,
p2 - p + 0.24 = 0
So,
Given:
p > 0.5.
So, take positive sign.
We get:
So,
Answer is:
0.6
(b)
By Propertyof Bernoulli Distribution:
E(X) = p
Thus, we get:
E(X) = p = 0.6
So,
Answer is:
0.6
If X follows Bernoulli distribution Bp,p > 0.5 and V(X) 0.24 . E(X)?
The 2 guesses are incorrect. Please help to solve correct answer 0.24, calculate the following: If X follows Bernoulli distribution B, p > 0.5 and V(X) .48 .48 ELX] 48
3. Let X N(20,1). What is P(X > 20) ? a) 0.25 b) 0.5 c) 0.75 d) 0.99
Exercise 6.14 Let y be distributed Bernoulli P(y = 1) unknown 0<p<1 p and P(y = 0) = 1-p f or Some (a) Show that p E( (b) Write down the natural moment estimator p of . (c) Find var (p) (d) Find the asymptotic distribution of vn (-p) as no. as n> OO.
Let random variable X follows an exponential distribution with probability density function fx (2) = 0.5 exp(-x/2), x > 0. Suppose that {X1, ..., X81} is i.i.d random sample from distribution of X. Approximate the probability of P(X1+...+X81 > 170). A. 0.67 B. 0.16 C. 0.33 D. 0.95 E. none of the preceding
Suppose that a random variable X follows N(45,12). What is P(X > 39)? Select one: O a. 0.6915 O b. 0.3085 c. 0.5
Let X, Y ~ 10,11 independently. Find P(max(X, Y} > 0.8 1 min(X, Y} = 0.5)
.X, be an id sample from the distribution r >1 (a) Using this distribution, find Eflog(X) 0.
Let X ~ Geomeric(p). Using Chebyshev's inequality find an upper bound for P(|X – E[X]] >b).
Modify X and apply Markov's inequality to upper bound P(X > 3) when X > 2 and E[X] = 2.5.
Look at the image, thank you. For a standard normal distribution, find c if P(z>c) = 0.6906 c=