Question

From information on a previous question: The mean systolic blood pressure for a population of patients (µ) from a local clinic is 130 with a standard deviation (σ) of 18.

What is the probability that a patient chosen at random will have a systolic blood pressure between 93 and 126? Rounded to the nearest ten thousandth (4 places to the right of the decimal).

Answer 1

The following information has been provided:

u = 130, 0 = 18

We need to compute The corresponding z-values needed to be computed are:

Pr(93 < X < 126

2 = 1; -- _ 93 126 -20556

z) – 2 – + _ 126 – 130 = -0.222

Therefore, we get:

Pr(93 < X < 126) = Pr ( 937130 < z < 1207a 130 ) = Pr(-2.0556 < Z 5-0.2222) = Pr(Z < -0.2222) – Pr(Z < -2.0556) = 0.4121 – 0.0199 = 0.3922

The following is obtained graphically:

$normaldistributiongrapher.php?mean=130&s$

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