If X is a geometric random variable Geometric(p), i.e., P(X=k) = p (1-p)k, k=0,1,2,... then E(X) is 1/p0p(1/p)-1(1/p)+1N/A (Select One) (5) If X is a normal N(0,1)N(0,1) random variable then E(X) is 0011−1−1+∞+∞−∞−∞N/A (Select One) (6) Find the variance of XX when X∼Geometric(p)X∼Geometric(p).The correct answer is 1−pp21−pp21p21p21p1p1p−11p−1None of the aboveN/A **Problem (4):** If $ X $ is a geometric random variable $ \text{Geometric}(p) $, i.e., $ P(X=k) = p(1-p)^k $, for $ k=0,1,2,\ldots $, then $ E(X) $ is- $ \frac{1}{p} $ (Selected Option)- 0- $ p $- $ \frac{(1/p)-1}{1} $- $ \frac{(1/p)+1}{1} $- N/A**Problem (5):** If $ X $ is a normal $ N(0, 1) $ random variable then $ E(X) $ is- 0 (Selected Option)- 1- $-1$- $+\infty$- $-\infty$- N/A**Problem (6):** Find the variance of $ X $ when $ X \sim \text{Geometric}(p) $.The correct answer is- $ \frac{1-p}{p^2} $ (Selected Option)- $ \frac{1}{p^2} $- $ \frac{1}{p} $- $ \frac{1}{p}-1 $- None of the above- N/A

Answered: (4) IfXis a geometric random variable…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

If X is a geometric random variable Geometric(p), i.e., P(X=k) = p (1-p)^k, k=0,1,2,... then E(X) is

		1/p	0	p	(1/p)-1	(1/p)+1	N/A
	(Select One)

(5) If X is a normal N(0,1)N(0,1) random variable then E(X) is

		00	11	−1−1	+∞+∞	−∞−∞	N/A
	(Select One)

(6) Find the variance of XX when X∼Geometric(p)X∼Geometric(p).

The correct answer is

1−pp21−pp2

1p21p2

1p1p

1p−11p−1

None of the above

N/A

**Problem (4):** If $ X $ is a geometric random variable $ \text{Geometric}(p) $, i.e., $ P(X=k) = p(1-p)^k $, for $ k=0,1,2,\ldots $, then $ E(X) $ is

- $ \frac{1}{p} $ (Selected Option)
- 0
- $ p $
- $ \frac{(1/p)-1}{1} $
- $ \frac{(1/p)+1}{1} $
- N/A

**Problem (5):** If $ X $ is a normal $ N(0, 1) $ random variable then $ E(X) $ is

- 0 (Selected Option)
- 1
- $-1$
- $+\infty$
- $-\infty$
- N/A

**Problem (6):** Find the variance of $ X $ when $ X \sim \text{Geometric}(p) $.

The correct answer is

- $ \frac{1-p}{p^2} $ (Selected Option)
- $ \frac{1}{p^2} $
- $ \frac{1}{p} $
- $ \frac{1}{p}-1 $
- None of the above
- N/A

Expert Solution

Step 1

Geometric random variable represents the number of trials required to get first success that means the random variable x is the number of failures and on ${(x + 1)}^{t h}$ trial is a success. The probability mass function is given by $P (x) = p {(1 - p)}^{x}$ .

A normal distribution follows a bell shaped curve which is centred at the mean and it is also called Gaussian distribution. The normal distribution with 0 mean and 1 standard deviation is called standard normal distribution.

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(4) IfXis a geometric random variable Geometric(p), ie., PX=k) = p (1-p)Š, k=0,1,2,... then EX) is 1/p (1/p)-1 (1/p)+1 N/A (Select One)