A little help? Thanks

For the following discrete random variable X with probability distribution:

Black_prince [1.1K]

Answer:

(a) The probability distribution is shown in the attachment.

(b) The value of E (Y) is 7.85.

(c) The value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d) The value of P (Y ≤ 2) is 0.60.

(e) Verified that the value of E (Y) is 7.85.

Step-by-step explanation:

(a)

The random variable Y is defined as: $Y=3X^{2}-2X+1$

For X = {0, 1, 2, 3} the value of Y are:

$X=0;\ Y=3\times(0)^{2}-2\times(0)+1 =1$

$X=1;\ Y=3\times(1)^{2}-2\times(1)+1 =2$

$X=2;\ Y=3\times(2)^{2}-2\times(2)+1 =9$

$X=3;\ Y=3\times(3)^{2}-2\times(3)+1 =22$

The probability of Y for different values are as follows:

P (Y = 1) = P (X = 0) = 0.20

P (Y = 2) = P (X = 1) = 0.40

P (Y = 9) = P (X = 2) = 0.15

P (Y = 22) = P (X = 3) = 0.25

The probability distribution of Y is shown below.

(b)

The expected value of a random variable using the probability distribution table is:

$E(U)=\sum[u\times P(U=u)]$

Compute the expected value of Y as follows:

$E(Y)=\sum [y\times P(Y=y)]\\=(1\times0.20)+(2\times0.40)+(9\times0.15)+(22\times0.25)\\=7.85$

Thus, the value of E (Y) is 7.85.

(c)

Compute the expected value of X as follows:

$E(X)=\sum [x\times P(X=x)]\\=(0\times0.20)+(1\times0.40)+(2\times0.15)+(3\times0.25)\\=1.45$

Compute the expected value of X² as follows:

$E(X^{2})=\sum [x^{2}\times P(X=x)]\\=(0^{2}\times0.20)+(1^{2}\times0.40)+(2^{2}\times0.15)+(3^{2}\times0.25)\\=3.25$

Thus, the value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d)

Compute the value of P (Y ≤ 2) as follows:

$P (Y\leq 2)=P(Y=1)+P(Y=2)=0.20+0.40=0.60$

Thus, the value of P (Y ≤ 2) is 0.60.

(e)

The value of E (Y) is 7.85.

$E(Y)=E(3X^{2}-2X+1)=3E(X^{2})-2E(X)+1$

Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).

$E(Y)=3E(X^{2})-2E(X)+1\\=(3\times 3.25)-(2\times1.45)+1\\=7.85$

Hence verified.

3 0

3 years ago

In a certain state car license plates are numbered in the form of LL NNNN, where L stands for a letter of the alphabet and N sta

givi [52]

Given:
LL NNNN

L - alphabet
N - number from 0 to 9

There repetition is allowed so. there are 26 choices in both Ls and 10 choices in the 4 Ns.

L L N N N N
26 * 26 * 10 * 10 * 10 * 10 = 6,760,000

There are 6,760,000 car license plate outcomes that are possible if all are allowed.

8 0

3 years ago

Which expression represents the difference of (8x-5y) and (2x-y)?

Alex

(8x-5y)-(2x-y)
8x-5y-2x+y
6x-4y

6 0

3 years ago

I need help with this math TEST SOMEONE HELP PLEASE

attashe74 [19]

Answer: b_oo_b pics

Step-by-step explanation:

get out ur phone and take off shirt and bra then look in mirror and take some real hot pics and send them to Donald trump