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 need to calculate the z-score first.
z = (X - μ) / σ
z = (103 - 120) / 7
z = -2.43
Now, we can use a Z-table or a calculator to find the probability corresponding to a z-score of -2.43. From the Z-table, we find that the probability of a z-score being less than -2.43 is approximately 0.0078.
Since we want to find the probability that the systolic blood pressure is higher than 103, we need to subtract this probability from 1:
P(X > 103) = 1 - P(X < 103)
P(X > 103) = 1 - 0.0078
P(X > 103) ≈ 0.992
Therefore, the probability that a randomly selected male's systolic blood pressure will be higher than 103 is approximately 0.992 to the nearest thousandth.