Answer:
a) [tex]\frac{8}{9}[/tex] =0.89
b) 0.8413
Step-by-step explanation:
we know that
[tex]P(║ x -\mu║ \leq k\sigma) \geq 1 -\frac{1}{k^2}[/tex]
[tex]so, \mu = 150, \sigma = 20[/tex]
[tex]k = \frac{210 - 150}{20} =\frac{150 - 90}{20} = 3[/tex]
[tex]P(║ x -150 ║ \leq 3\times 20) \geq 1 -\frac{1}{3^2} = \frac{8}{9}[/tex]
x - N(150, 20^2)
[tex]P(X \leq 170) = 0.8413447[/tex]
check
[tex]P norm = \frac{170 - 150}{20}= 0.8413447[/tex]