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7) If the systolic blood pressure is normally distributed in the population with a mean=120, SD=20, X~N(120, 20) what is the probability of a random person having systolic blood pressure of 130 or greater (hint: calculate the Z score, then find the probability in the right tail)

Respuesta :

Answer:

Step-by-step explanation:

Given that the systolic blood pressure is normally distributed in the population with a mean=120, SD=20.

the probability of a random person having systolic blood pressure of 130 or greater

Convert 130 to Z score as

[tex]z=\frac{130-120}{20} =0.5[/tex]

Required probability=

P(Z\geq  0.5) =

=0.5-0.1915

=0.3085

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