For males in a certain town, the systolic blood pressure is normally distributed with a mean of 120 and a standard deviation of 7 What is the probability that a randomly selected male's systolic blood pressure will be higher than 103, to the nearest thousandth?
To find the probability that a randomly selected male's systolic blood pressure will be higher than 103, we first need to standardize the value using the z-score formula:
z = (X - μ) / σ
where:
X = 103 (value we want to find the probability for)
μ = 120 (mean)
σ = 7 (standard deviation)
z = (103 - 120) / 7
z = -17 / 7
z = -2.429
Now, we need to find the probability that a z-score is greater than -2.429. We can use a standard normal distribution table or a calculator to find this probability.
Looking up the z-score of -2.429 in a standard normal distribution table, we find that the probability is approximately 0.9931.
Therefore, the probability that a randomly selected male's systolic blood pressure will be higher than 103 is approximately 1 - 0.9931 = 0.0069, or 0.007 when rounded to the nearest thousandth.