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ludmilkaskok [199]

2 years ago

8

Trey recorded the number of ounces of water he drank each day in the histogram below. Which of the following would be the best m

easure of variability?
a) mean
b) mean absolute deviation
c) median
d) interquartile range

2 answers:

hram777 [196]2 years ago

8 0

Mean absolute deviation

Anna11 [10]2 years ago

6 0

I believe it's mean absolute deviation

Step-by-step explanation:

You might be interested in

The vertices of quadrilateral MNPQ are M(−3,−2),N(−1,4),P(2,4), and Q(4,−2). Translate quadrilateral MNPQ using the vector ⟨3,−4

muminat

Answer:

see explanation

Step-by-step explanation:

A translation using the vector < 3, - 4 > , means

Add 3 to the x- coordinate and subtract 4 from the y- coordinate, so

M (- 3, - 2 ) → M' (- 3 + 3, - 2 - 4 ) → M' (0, - 6 )

N (- 1, 4 ) → N' (- 1 + 3, 4 - 4 ) → N' (2, 0 )

P (2, 4 ) → P' (2 + 3, 4 - 4 ) → P' (5, 0 )

Q (4, - 2 ) → Q' (4 + 3, - 2 - 4 ) → (7, - 6 )

4 0

2 years ago

Suppose a sample of 256 was taken from a population with a standard deviation of 64 centimeters. What is the margin of error for

Ganezh [65]

I think the answer is b

8 0

3 years ago

Read 2 more answers

I really don't know how to do this

Vaselesa [24]

When it comes to finding points on a graph, think of the saying, "you have to learn how to walk before you climb."

7 0

3 years ago

40 is what percent of 32?

Daniel [21]

Answer:

125%

Step-by-step explanation:

Is means equals and of means multiply

40 = P * 32 where P is in decimal form

Divide each side by 32

40/32 = P

1.25 = P

Now change to percent form ( multiply by 100)

125%

7 0

3 years ago

Read 2 more answers

Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste

lidiya [134]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$ In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$ And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

Step-by-step explanation:

Previous concepts

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).

Solution to the problem

For this case we have the following distribution given:

X 3 4 5 6

P(X) 0.07 0.4 0.25 0.28

We can calculate the mean with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74$

In order to find the variance we need to calculate first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36$

And the variance is given by:

$Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924$

And the deviation would be:

$Sd(X) =\sqrt{0.8924} =0.9447$

3 0

3 years ago