Helppp please, cause I really need it tomorrow

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

A baseball player was that bat 473 times during a regular season. if he made a hit 31. 5 % of the time he was at bat how many hi

alex41 [277]

Well, to calculate that, you have to multiply the number of bats by the "number" version of 31.5 %, which once you multiply them together, it comes out to be about 149 (148.995)

3 0

2 years ago

Read 2 more answers

Find the amount of interest on $1,500 invested for 20 days at 6.75%, compounded daily. $_____

LekaFEV [45]

Answer:

The required amount of interest is $5538.

Step-by-step explanation:

Given : $1,500 invested for 20 days at 6.75%, compounded daily.

To find : The amount of interest ?

Solution :

Applying compound interest formula,

$A=P(1+r)^t$

Where, A is the amount

P is the principal P=$1500

r is the interest rate r=6.75%=0.0675 compounded daily

t is the time t=20 days

Substitute the value in the formula,

$A=P(1+r)^t$

$A=1500(1+0.0675)^{20}$

$A=1500(1.0675)^{20}$

$A=1500\times 3.69$

$A=\$5538$

Therefore, The required amount of interest is $5538.

4 0

3 years ago

Find the coordinates of the midpoint of a segment with the endpoints (12,2) and (7,9)

olga55 [171]

Answer:

(9 $\frac{1}{2}$ , 5 $\frac{1}{2}$ ) or (9 1/2, 5 1/2) (same things, just typed two different ways)

Step-by-step explanation

I can't remember the midpoint formula so I'll use a different method to get the midpoint.

since the ratio of a midpoint is 1:1 and that as a fraction is 1/2, we can take these coordinates and designate which one is first and which one is second (12, 2) will be my second and (7,9) will be my first. The formula I will use will be $\frac{m}{m+n} (second X, - first X) + first x$ , so 1/2( 12 - 7) +7. now distribute the 1/2 to both the numbers $\frac{1x12}{2} -\frac{1x7}{2} + 7 \\ \\ 12/2 -7/2 = 5/2 + 7 = 9.5$ or $9\frac{1}{2}$ . The same works with the Y values, which would equal up to 5.5 or 5 $\frac{1}{2}$