Answer:
0.121
Step-by-step explanation:
From the given information:
Let X be a hypergeometric random variable. i.e X [tex]\sim[/tex] h (x;:6,7,12)
To determine the mean and the standard deviation of X of a hypergeometric random variable; we have:
[tex]\mu_x = E(X)[/tex]
[tex]\mu_x = 6 \times \dfrac{7}{12}[/tex]
[tex]\mu_x =1 \times \dfrac{7}{2}[/tex]
[tex]\mu_x =3.5[/tex]
The standard deviation is :
[tex]\sigma _x = \sqrt{(\dfrac{12-6}{12-1}) \times 6 \times (\dfrac{7}{12})\times (1-\dfrac{7}{12})}[/tex]
[tex]\sigma _x = \sqrt{(\dfrac{6}{11}) \times 6 \times (\dfrac{7}{12})\times (\dfrac{5}{12})}[/tex]
[tex]\sigma _x = \sqrt{\dfrac{35}{44}}[/tex]
[tex]\sigma _x = 0.892[/tex]
However, the inequality for the event showcasing how X exceeds its mean value by more than 1 standard deviation is :
[tex]X \geq \mu_x +\sigma_x[/tex]
[tex]X \geq 3.5 +0.892[/tex]
[tex]X \geq 4.392[/tex]
[tex]X \geq 5[/tex]
∴
[tex]P(X \geq \mu_x + \sigma_x ) = P(X \geq 5)[/tex]
[tex]P(X \geq \mu_x + \sigma_x ) = 0.121[/tex]
