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4 years ago

What conclusion can be derived by comparing the central tendencies of the two data sets?

Mathematics

1 answer:

Vika [28.1K]4 years ago

6 0

Answer:

See below.

Step-by-step explanation:

The mean of set A = 4.3625

The mean of set B = 4.1.

The means are very close.

The median of set A = 4.5

The median of set B = 3.5.

The mean and median of Set A are quite close so we can say that the data is close to Symmetrical.

In the case of set B the mean and median are further apart so the data is more skewed . it is skewed towards the lower values (to the left).

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Graph the six terms of a finite series where a1 = −3 and r = 1.5.

Pani-rosa [81]

Answer:According to the site that i have tested on, the correct answer is d

Step-by-step explanation: Using a graphing calculator, it can be seen that those are the first six terms that are seen.

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3 years ago

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What is the range of the given set of ordered pairs?<br> (9,-2) (4,3) ( 8, 10) (-4, 8)

otez555 [7]

Answer:

(-2,3,10,8)

Step-by-step explanation:

(x,y)=(domain,range)

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3 years ago

Let Y1 and Y2 denote the proportion of time during which employees I and II actually performed their assigned tasks during a wor

Lemur [1.5K]

Answer:

Step-by-step explanation:

From the information given:

The joint density of $y_1$ and $y_2$ is given by:

$f_{(y_1,y_2)} \left \{ {{y_1+y_2, \ \ 0\ \le \ y_1 \ \le 1 , \ \ 0 \ \ \le y_2 \ \ \le 1} \atop {0, \ \ \ elsewhere \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \right.$

a)To find the marginal density of $y_1$ .

$f_{y_1} (y_1) = \int \limits ^{\infty}_{-\infty} f_{y_1,y_2} (y_1 >y_2) \ dy_2$

$=\int \limits ^{1}_{0}(y_1+y_2)\ dy_2$

$=\int \limits ^{1}_{0} \ \ y_1dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1 \ \int \limits ^{1}_{0} dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1[y_2]^1_0 + \bigg [ \dfrac{y_2^2}{2}\bigg]^1_0$

$= y_1 [1] + [\dfrac{1}{2}]$

$= y_1 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_1)}= \left \{ {{y_1+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

The marginal density of $y_2$ is:

$f_{y_1} (y_2) = \int \limits ^{\infty}_{-\infty} fy_1y_1(y_1-y_2) dy_1$

$= \int \limits ^1_0 \ y_1 dy_1 + y_2 \int \limits ^1_0 dy_1$

$=\bigg[ \dfrac{y_1^2}{2} \bigg]^1_0 + y_2 [y_1]^1_0$

$= [ \dfrac{1}{2}] + y_2 [1]$

$= y_2 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_2)}= \left \{ {{y_2+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

$P\bigg[y_1 \ge \dfrac{1}{2}\bigg |y_2 \ge \dfrac{1}{2} \bigg] = \dfrac{P\bigg [y_1 \ge \dfrac{1}{2} . y_2 \ge\dfrac{1}{2} \bigg]}{P\bigg[ y_2 \ge \dfrac{1}{2}\bigg]}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} f_{y_1,y_1(y_1-y_2) dy_1dy_2}}{\int \limits ^1_{\frac{1}{2}} fy_1 (y_2) \ dy_2}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} (y_1+y_2) \ dy_1 dy_2}{\int \limits ^1_{\frac{1}{2}} (y_2 + \dfrac{1}{2}) \ dy_2}$