Question

Systolic blood pressure for a group of women is normally distributed, with a mean of 121 and a standard deviation of 9. Find the probability that a woman selected at random has the following blood pressures. (Round your answers to four decimal places.) (a) greater than 136 (b) less than 114 (c) between 114 and 128

Answer 1

the probability that a woman selected at random has a blood pressure between 114 and 128 is 0.5588.

What is a Z-table?

A z-table also known as the standard normal distribution table, helps us to know the percentage of values that are below (or to the left of the Distribution) a z-score in the standard normal distribution.

Given the mean is 121 while the standard deviation of the women is 9. Therefore, Using the z-table, the probability can be found.

(a) The probability that a woman selected at random has blood pressures greater than 136.

$P(x > 136) = 1 - P(x < 136)\\\\P(x > 136) = 1 - P(z < \dfrac{x-\mu}{\sigma})$

                  $=1 - P(z < \dfrac{136-121}{9})\\\\=1 - P(z < 1.667)\\\\=1-0.9515\\\\=0.0485$

(b) The probability that a woman selected at random has a blood pressure less than 114.

$P(x < 114)= P(z < \dfrac{114-121}{9})$

                  $= P(z < -0.77)\\\\= 0.2206$

(c) The probability that a woman selected at random has a blood pressure between 114 and 128.

$P(114 < x < 128)= P(\dfrac{114-121}{9} < z < \dfrac{128-121}{9})$

                  $= P(-0.77 < z < 0.77)\\\\= P(z < 0.77)-P(z < -0.77)\\\\= 0.7794 - 0.2206\\\\=0.5588$

Hence, the probability that a woman selected at random has a blood pressure between 114 and 128 is 0.5588.

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