Which of the following equations cannot be solved using real numbers?

- To check the legitimacy of a continuous probability density function the area under the curve over the domain must be equal to 1. In other words the following:

$\int\limits^a_b {f_1( x )} \, dx = 1\\\\ \int\limits^a_b {f_2( x )} \, dx = 1\\$

- We will perform integration of each given pdf as follows:

$\int\limits^a_b {f_1(x)} \, dx = \int\limits^2_0 {0.5(3x - x^3 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - 0.125x^4 ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - 0.125*(16) ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 2 ] = 1\\$

$\int\limits^a_b {f_2(x)} \, dx = \int\limits^2_0 {0.5(3x - x^2 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - \frac{x^3}{6} ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - \frac{(8)}{6} ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 1.3333 ] = 1.67 \neq 1 \\$

Answer: f1 ( x ) is a valid pdf; however, f2 ( x ) is not a valid pdf.

B)

- A random variable ( X ) denotes the resistance of a randomly chosen resistor, and the pdf is given as follows:

$f ( x ) = kx$ if 8 ≤ x ≤ 10

0 otherwise.

- To determine the value of ( k ) we will impose the condition of validity of a probability function as follows:

$\int\limits^a_b {f(x)} \, dx = 1\\$

- Evaluate the integral as follows:

$\int\limits^1_8 {kx} \, dx = 1\\\\\frac{kx^2}{2} ]\limits^1^0_8 = 1\\\\k* [ 10^2 - 8^2 ] = 2\\\\k = \frac{2}{36} = \frac{1}{18}$ ... Answer

- To determine the CDF of the given probability distribution we will integrate the pdf from the initial point ( 8 ) to a respective value ( x ) as follows:

$cdf = F ( x ) = \int\limits^x_8 {f(x)} \, dx\\\\F ( x ) = \int\limits^x_8 {\frac{x}{18} } \, dx\\\\ F ( x ) = [ \frac{x^2}{36} ] \limits^x_8\\\\F ( x ) = \frac{x^2 - 64}{36}$

To determine the probability p ( 8.6 ≤ x ≤ 9.8 ) we will utilize the cdf as follows:

p ( 8.6 ≤ x ≤ 9.8 ) = F ( 9.8 ) - F ( 8.6 )

p ( 8.6 ≤ x ≤ 9.8 ) = $\frac{(9.8)^2 - 64}{36} - \frac{(8.6)^2 - 64}{36} = 0.61333$

ii) To determine the conditional probability we will utilize the basic formula as follows:

p ( x ≤ 9.8 | x ≥ 8.6 ) = p ( 8.6 ≤ x ≤ 9.8 ) / p ( x ≥ 8.6 )

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 1 - p ( x ≤ 8.6 ) ]

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 1 - 0.27666 ]

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 0.72333 ]

p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.84792 ... answer

3 0

3 years ago

Miss Uche made a fresh pot of coffee in the morning. By the end of the day only 3 cups remained. If 85% of the coffee had been c

Pavlova-9 [17]

Answer: She made about 3.5 cups of coffee.

Step-by-step explanation:

n * 85% = 3

n= 3/0.85= 3.529

3.5 * 0.85= 2.975

6 0

2 years ago