All categories

2 years ago

What is the mean absolute deviation for the data set

Mathematics

1 answer:

motikmotik2 years ago

8 0

Answer:

Mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean.

Step-by-step explanation:

You might be interested in

How to solve for area of triangle where A=67 degrees, b=11yd, and c=24yd

skad [1K]

Use the formula of an area of a tringle:

$A=\dfrac{1}{2}bc\sin A$

We have b = 11 yd, c = 24 yd and m∠A = 67°

$A=\dfrac{1}{2}(11)(24)\sin67^o$

$\sin67^o\approx0.9205\\\A=(11)(12)(0.9205)=121.506\ yd^2$

4 0

3 years ago

Add the two mixed numbers 1 1/3 + 3 2/3 = ?

pychu [463]

im not sureeeeeeeeeeeeeeeeeeeeeeeee

8 0

2 years ago

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Rom4ik [11]

Answer:

Step-by-step explanation:

100- 51 = 49

7 0

3 years ago

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

The Confidence interval is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and the critical values are:

$\chi^2_{\alpha/2}=20.09$

$\chi^2_{1- \alpha/2}=1.65$

And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0

3 years ago

(3x + 3) (2x − 2) Please help me

kupik [55]

The answer is 6(x^2 -1) I have provided a picture with my work and the answer

8 0

3 years ago