For women aged 19-24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. Find the probability that 23 randomly selected women aged 19-24 have a mean systolic blood pressure greater than 121.5. Round your answer to 4 decimal places.

Answer: p = 0.00714

Step-by-step explanation:

Since the population standard deviation of the distribution is known, the z test (which produces the z score) is the perfect test for finding the probability of the data set in the question.

From the question,

Population mean (u) = 114.8

Sample mean (x) = 121.5

Population standard deviation (σ) = 13.1

Sample size (n) = 23

The z score formulae is given below as

Z = x - u/σ/√n

Z = 121.5 - 114.8 / (13.1/√23)

Z = 6.7 / (13.1/√23)

Z = 6.7/2.731

Z = 2.45.

The question is interested in knowing the probability of mean systolic blood pressure greater than 121.5.

This implies that we are looking for the probability at which our z score is greater than 2.45: P (z>2.45)

The z score (z=2.45) has divided the distribution into two regions, z2.45 (area to the right of distribution).

Hence, p (z>2.45) + p (z<2.45) = 1

To get a the probability, we have to use a standard normal distribution table.

The table we have here gives probability of the distribution to the left (that's area towards the left), hence we need to find p (z<2.45) first

From the table p (z<2.45) = 0.99286

But p (z>2.45) = 1 - p (z<2.45)

p (z>2.45) = 1 - 0.99286

p (z>2.45) = 0.00714