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hammer [34]
3 years ago
5

The ________ can be a misleading measure of central tendency if the distribution of scores is ________ . A.mean; B.skewed median

; C.skewed mean; D. normal median; E.normal
Mathematics
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

C) Mean, Skewed.

Step-by-step explanation:

The mean is easily influenced by outliers in a distribution. So if a Distribution of scores is skewed, then it will mislead the readers. In a skew distribution, either positive or negative, the Median will turn to be the central value. Usually, the mean is used besides other statistical parameters.

In a Normal distribution, the mean is right in the middle, symmetrically dividing the "bell shape" curve. The Mean, in this case, coincides with the mode, and the Median.

On the other hand, the Median is way more trustable central tendency measure, more resistant to outliers. So in a Skewed distribution, a Median is preferred over the Mean and use alongside the Mode.

You might be interested in
I need help fast ....​
Cloud [144]
Since x=-10, plug that into the equation.

f(x) = 2(-10) + 11
= -20 + 11
y = -9

So the ordered pair in (x,y) terms is (-10,-9).
4 0
3 years ago
A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2
Yuliya22 [10]

A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2 and 5.1.

Answer: The correct option is B) about 34%

Proof:

We have to find P(4.2

To find P(4.2, we need to use z score formula:

When x = 4.2, we have:

z = \frac{x-\mu}{\sigma}

          =\frac{4.2-5.1}{0.9}=\frac{-0.9}{0.9}=-1

When x = 5.1, we have:

z = \frac{x-\mu}{\sigma}

          =\frac{5.1-5.1}{0.9}=0

Therefore, we have to find P(-1

Using the standard normal table, we have:

P(-1= P(z

                               =0.50-0.1587

                               =0.3413 or 34.13%

                               = 34% approximately

Therefore, the percent of data between 4.2 and 5.1 is about 34%

7 0
3 years ago
It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir
Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, we find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

M = z*\frac{\sigma}{\sqrt{n}}

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

n = 74.4

We need a sample size of at least 75.

6 0
3 years ago
Read 2 more answers
Myra is evaluating the expression –31.7 + 4.5x, when x = 2.1.
mylen [45]

Answer:

-22.25

Step-by-step explanation:

-31.7+4.5(2.1)

-31.7+9.45

-22.25

8 0
3 years ago
Read 2 more answers
La señora esta caminando en la calle y saludo a todos 
grandymaker [24]
Hola, necesita ayuda con translation? 
La señora esta caminando en la calle y Saludo a todos.
The woman is walking in the street and greets everyone.
5 0
3 years ago
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