A random variable X follows a Binomial distribution; i.E. X ~ B(12, p) with a variance of 1.92. Compute the possible values of p

Mathematics

1 answer:

frutty [35]3 years ago

4 0

<h3>2 Answers: p = 0.2 and p = 0.8</h3>

==================================================

Explanation:

Any binomial distribution has two unique components to it: The number of trials n, and the probability of success p.

We're given n = 12, and we need to find p.

The variance of a binomial probability distribution is given by the formula n*p*(1-p). Plug in n = 12 and set this equal to 1.92

So the equation to solve is 12p*(1-p) = 1.92

Solve for p using a graphing calculator or you could apply the quadratic formula. Either way, you should get the two results of p = 0.2 and p = 0.8

Note that if p = 0.2, then 1-p = 1-0.2 = 0.8, and things flip around when p = 0.8.

You might be interested in

A 160lb person has 4.6 quarts of blood what is the direct variation

zhenek [66]

So weight=x
blood=y

we got that when
y=kx
when y=4.6, then x=160
4.6=160k
divide both sides by 160
0.02875=k

the direct variation is
y=0.02875x

so plug in x and get y

the answers in (x,y) form is
(0,0)
(50,1.4375)
(75,2.15625)
(100,2.875)
(125,3.59375)
(150,4.3125)

5 0

3 years ago

Tell whether the line or segment is best described as a radius, chord, diameter, secant, or tangent of ⨀O.

PilotLPTM [1.2K]

BC = TAN

AB = SEC

OB = RADIUS

DB= diameter

third AB is = cord

5 0

3 years ago

Hexagon DEFGHI is translated 8 units down and 3 units to the right. If the coordinates of the pre-image of point F are

Zarrin [17]

The coordinates of F' is (-9 + 3, 2 - 8) = (-6, -6)

6 0

4 years ago

Read 2 more answers

There's a width of 8 in., a length of 20 in., and a height of 12 in.

Setler [38]

Answer:

Approximately $24.7\; \rm in$ .

While there are three diagonals in a box (a rectangular prism,) all three diagonals goes through the same point- the centroid of this box.

For a maximum-length poster to fit in this box, it would have to be on one of the main diagonals of this box. Hence, any maximum-length poster that fits in this box would go through the centroid of this box.

It's not possible to force more than one posters to go through the same point (i.e., the centroid) in space. Hence, it would not be possible to fit a second maximum-length poster into this box.

This argument does not apply to $21.5\; \rm in$ posters. These posters are shorter than the diagonal of this box; they could fit inside the box without having to go through a particular point in space.

Step-by-step explanation:

The longest poster that could be fit into this box (a rectangular prism) would be as long as the longest line segment in this box. That line segment would be one of the three diagonals of this box.

Apply the Pythagorean theorem twice to find the length of that diagonal.

Start by finding calculating the diagonal of the base of this box. The base of this box is a rectangle with width $8\; \rm in$ and length $10\; \rm in$ . The length of its diagonal would be $\sqrt{8^2 + 10^2}$ inches.