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2 years ago

Explain when the median of a data set is a better measure of center than the mean

Mathematics

2 answers:

choli [55]2 years ago

7 0

Answer:

Sample Response: When there is an outlier in the data set, the dot plot or histogram will be skewed. In a skewed representation, the mean is pulled up or down toward the tail of the data. Therefore, skewed data affects the mean more than the median.

What did you include in your response? Check all that apply.

The dot plot or histogram will be skewed.

The mean is pulled up or down toward the tail.

The mean is affected more than the median.

vlada-n [284]2 years ago

5 0

The dot plot or histogram will be skewed.

The mean is pulled up or down toward the tail.

The mean is affected more than the median.

You might be interested in

In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline

VLD [36.1K]

Answer:

<u>The measure of the arc CD = 64°</u>

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

So, as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

<u>∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°</u>

6 0

1 year ago

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution

oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

$P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!$

Then, we can calculate P(X ≤ 5) as:

$P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999$

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

$P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080$

4 0

2 years ago

According to a 2014 research study of national student engagement in the U.S., the average college student spends 17 hours per w

Nadusha1986 [10]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 17

For the alternative hypothesis,

µ < 17

This is a left tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 80,

Degrees of freedom, df = n - 1 = 80 - 1 = 79

t = (x - µ)/(s/√n)

Where

x = sample mean = 15.6

µ = population mean = 17

s = samples standard deviation = 4.5

t = (15.6 - 17)/(4.5/√80) = - 2.78

We would determine the p value using the t test calculator. It becomes

p = 0.0034

Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.

The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.

4 0

2 years ago

Complete the statements for the graph of f(x) = |x|. The domain of the function is . The range of the function is . The graph is

ioda

Answer:

Domain of f(x): All real numbers

Range of f(x): $f(x) \geq0$ ---> [0, ∞)

The graph is over the interval (0, ∞)

Step-by-step explanation:

The function absolute value of x that is written: f(x) = | x | assigns only positive values to f(x) at any value of x.

For example, if for f(x) = | x | we make

f(-23) = | -23 | then the absolute value of x "transforms" the number -23 to +23.

Then f(-23) = | -23 | = 23

This means that f(x) will be positive for any value of x.

Therefore, we can conclude that:

Domain of f(x): All real numbers

Range of f(x): $f(x) \geq0$ ---> [0, ∞)

The graph is over the interval (0, ∞)

5 0

2 years ago

Read 2 more answers

A local museum charges $25 per adult and $12 per child for admission fees. At the end of a day, the museum made $9,014 in total

Natali [406]

Answer:

There were 172 guests that were children and 278 guests that were adults

Step-by-step explanation:

Step 1: State what is known

It is $25 per adult

It is $12 per child

The made $9014 dollars

There were 450 guests

Step 2: Define equations

25y + 12x = 9014 -----1

x + y = 450 ------------2

Step 3: Rearrange equation 2 for x

x + y = 450

x = 450 - y --------------3

Step 4: Substitute 3 into 1 for y and solve for y

25y + 12(450 - y) = 9014

25y + 5400 - 12y = 9014

13y = 3614

y = 3614/13

y = 278

Step 5: Substitute y = 278 into 3 to solve for x

x = 450 - (278)

x = 172

Therefore 172 children and 278 adults visited the museum

4 0

2 years ago