Ask question

Ask question

All categories

3 years ago

15

If the outliers are not included, what is the mean of the data set

1 answer:

Naddika [18.5K]3 years ago

5 0

804÷12
=67
so the mean is 67

You might be interested in

Which measure of central tendency is the most affected by outliers

aliina [53]

Mean measure of central tendency is the most affected by outliers. Then the correct option is B.

<h3>What is Mean?</h3>

The mean is the straightforward meaning of the normal of a lot of numbers. In measurements, one of the markers of focal propensity is the mean. The normal is alluded to as the number-crunching mean. It's the proportion of the number of genuine perceptions to the absolute number of perceptions.

Mean measure of central tendency is the most affected by outliers.

Then the correct option is B.

More about the mean link is given below.

brainly.com/question/521501

#SPJ1

7 0

2 years ago

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

Evaluate 3(a + b + c) 2 for a = 2, b = 3, and c = 4.

garri49 [273]

Answer:

54

Step-by-step explanation:

Given:

$3(a + b + c) 2$

Input the numbers provided into the equation

3(2 + 3 + 4)2

Add all of the numbers inside the parentheses

3(9)2

Multiply

3 * 2 = 6

6 * 9 = 54

6 0

3 years ago

Read 2 more answers

An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholde

kherson [118]

Answer:

Step-by-step explanation:

Given that X is the number of months between successive payments

Cumulative distribution function of X is

$F(x) =0, x$

a) PMF of x would be

$P(x) =0, x$

b) $P(3 ≤ X ≤ 6) =0.86-0.44=0.42\\ P(4 ≤ X)=1-0.44=0.56\\P(3 < X < 6) =0.6\\P(4 < X) =1-0.48=0.52$

4 0

2 years ago

julianna wrote -16+ x = -8 on the board.Do not solve.Tell how you would simplify julianna's expression?

LUCKY_DIMON [66]

Answer:

x = 8

Step-by-step explanation:

→ Add 16 to both sides

x = 8

6 0

2 years ago