I am really stuck please help meth

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

6 0

3 years ago

What is the range of the function y = -x2 + 1?

meriva

<span><span>y = 2 + 2sec(2x) The upper part of the range will be when the secant has the smallest positive value up to infinity. The smallest positive value of the secant is 1 So the minimum of the upper part of the range of y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4 So the upper part of the range is [4, ) The lower part of the range will be from negative infinity up to when the secant has the largest negative value. The largest negative value of the secant is -1 So the maximum of the lower part of the range of y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0 So the lower part of the range is (, 0]. Therefore the range is (, 0] U [4, ) </span>
</span>

3 0

3 years ago

Finding Account Balances In Exercise, complete the table to determine the balance A for P dollars Invested at rate for years com

Arlecino [84]

Answer:

i have no idea

Step-by-step explanation:

4 0

3 years ago

Work out the perimeter of a quarter-circle with radius 7cm.

Flura [38]

Answer:

25

Step-by-step explanation:

p=1÷4pir+2r

1÷4×22÷7×7+2(7)

=11cm+14cm

=25

7 0

3 years ago

Read 2 more answers

Steven wants to buy a new video game. He needs at least $48. He has $26.00 saved. His brother said he would pay him $5 an hour t

Law Incorporation [45]

I think that the answer is C

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2 years ago