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dontcareanyonemo

23.11.2020 •

Mathematics

Help please :)<3 thx

Help please :)<3 thx

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Ответ:

kiannasmith46

08.03.2020

$(a)\ c = \frac{1}{15}$

$(b)\ 40\%$

$(c)\ 60\%$

Step-by-step explanation:

The given parameters can be represented as:

$P_Y(y) \ge 0, y =1,2,3,4,5$

P_y(y) = cy, y=1,2,3,4,5

Solving (a): The value of c

To do this, we make use of the following rule;

$\sum\limit^5_{y=1}P_Y(y_i) = 1$

Given that:

P_y(y) = cy, y=1,2,3,4,5

This is translated to:

c*1 + c * 2 + c * 3 + c * 4 + c * 5 = 1

c + 2c + 3c + 4c + 5c = 1

15c = 1

Solve for c

$c = \frac{1}{15}$

(b) The proportions of applications that requires at most 3 forms

This implies that: y = 1,2,3

So, we make use of:

$P(Y \le 3) = P(Y=1) + P(y=2) + P(Y=3)$

Recall that:

P_y(y) = cy, y=1,2,3,4,5

Substitute $c = \frac{1}{15}$

$P_y(y) =\frac{1}{15}y$

So:

$P(Y \le 3) = P(Y=1) + P(y=2) + P(Y=3)$

$P(Y\le 3) = \frac{1}{15} * 1 +\frac{1}{15} * 2 +\frac{1}{15} * 3$

$P(Y\le 3) = \frac{1}{15} +\frac{2}{15} +\frac{3}{15}$

Take LCM

$P(Y\le 3) = \frac{1+2+3}{15}$

$P(Y\le 3) = \frac{6}{15}$

$P(Y\le 3) = 0.4$

Express as percentage

$P(Y\le 3) = 0.4*100\%$

$P(Y\le 3) = 40\%$

(c) The proportions of applications that requires between 2 and 4 forms (inclusive)

This implies that: y = 2,3,4

So, we make use of:

$P(2 \le Y \le 4) = P(Y=2) + P(Y=3) + P(Y=4)$

$P(2 \le Y \le 4) = 2 * \frac{1}{15} + 3 * \frac{1}{15} + 4 * \frac{1}{15}$

$P(2 \le Y \le 4) = \frac{2}{15} + \frac{3}{15} + \frac{4}{15}$

Take LCM

$P(2 \le Y \le 4) = \frac{2+3+4}{15}$

$P(2 \le Y \le 4) = \frac{9}{15}$

$P(2 \le Y \le 4) = 0.6$

Express as percentage

$P(2 \le Y \le 4) = 0.6 * 100\%$

$P(2 \le Y \le 4) = 60\%$

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