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2 years ago

Please what is the interquartile range ?

Mathematics

1 answer:

AleksAgata [21]2 years ago

6 0

Answer:

Step-by-step explanation:

15 - 12 = 3.

The ends of the box are the lower and upper quartiles.

The difference upper - lower = the interquartile range.

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Let X be a Bernoulli rv with pmf as in Example 3.18. a. Compute E(X2 ). b. Show that V(X) 5 p(1 2 p). c. Compute E(X79).

spayn [35]

The Bernoulli distribution is a distribution whose random variable can only take 0 or 1

The value of E(x2) is p
The value of V(x) is p(1 - p)
The value of E(x79) is p

<h3>How to compute E(x2)</h3>

The distribution is given as:

p(0) = 1 - p

p(1) = p

The expected value of x2, E(x2) is calculated as:

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * (1- p) + 1^2 * p$

Evaluate the exponents

$E(x^2) = 0 * (1- p) + 1 * p$

Multiply

$E(x^2) = 0 +p$

Add

$E(x^2) = p$

Hence, the value of E(x2) is p

<h3>How to compute V(x)</h3>

This is calculated as:

$V(x) = E(x^2) - (E(x))^2$

Start by calculating E(x) using:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * (1- p) + 1 * p$

$E(x) = p$

Recall that:

$V(x) = E(x^2) - (E(x))^2$

So, we have:

$V(x) = p - p^2$

Factor out p

$V(x) = p(1 - p)$

Hence, the value of V(x) is p(1 - p)

<h3>How to compute E(x79)</h3>

The expected value of x79, E(x79) is calculated as:

$E(x^{79}) = \sum x^{79} * P(x)$

So, we have:

$E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p$

Evaluate the exponents

$E(x^{79}) = 0 * (1- p) + 1 * p$

Multiply

$E(x^{79}) = 0 + p$

Add

$E(x^{79}) = p$

Hence, the value of E(x79) is p

Read more about probability distribution at:

brainly.com/question/15246027

5 0

2 years ago

Solve for the ? angle

daser333 [38]

Answer:

Step-by-step explanation:

since all angles add up to 180

and since this is a right triangle the square in the corner is 90 degrees

so 40+90+x=180

130+x=180

subtract 130 from both sides

130-130+x=180-130

x=50

Hope that helps :)

7 0

3 years ago

What is 12+ 12-4 %2<br> Hhhhhhhhhhhhhhhhhhhhhhhhhhh

prohojiy [21]

Can you show me a picture of the problem?

6 0

2 years ago

If you had a billion dollars and you waste half of your money how much would you Have plzz heart this

Ne4ueva [31]

You would have half a billion

4 0

2 years ago

Read 2 more answers

At 3pm, the length of the shadow of a thin vertical pole standing on level ground is the same as the height of the pole. A while

aliya0001 [1]

Answer: The height of the pole is 175.97 ( approx )

Step-by-step explanation:

Let the height of the pole is x cm,

Also, let the angle of elevation of the sun to the pole at 3 pm is $\theta$

Thus, by the question,

$tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}$

$\implies tan\theta = \frac{x}{x}$ ( at 3pm the height of shadow = height of the pole)

$\implies tan\theta = 1$

$\implies \theta = 45^{\circ}$

Again, according to the question,

When the angle of elevation is $(\theta - 12^{\circ})$ ,

The height of shadow = x + 95,

$\implies tan(45-12)^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x$

$\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}$

$\implies -0.3505924068 x = - 61.6937213538$

$x=175.969930202\approx 175.97\text{ cm}$

4 0

3 years ago