Question

Suppose that the systolic blood pressure of healthy adults is normally distributed with mean μ=112 mmHg and standard deviation σ=10mmHg. What is the probability (as a decimal) that a randomly sampled healthy adult has systolic blood pressure less than 96mmHg ?

Answer 1

To find the probability that a randomly sampled healthy adult has systolic blood pressure less than 96mmHg, we need to calculate the z-score and then use the standard normal distribution table.

Step-by-step explanation:

To find the probability that a randomly sampled healthy adult has systolic blood pressure less than 96mmHg, we need to calculate the z-score and then use the standard normal distribution table.

First, we calculate the z-score using the formula z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation. For this case, x = 96mmHg, μ = 112mmHg, and σ = 10mmHg.

Substituting the values into the formula, we have: z = (96 - 112) / 10 = -1.6.

Next, we look up the z-score -1.6 in the standard normal distribution table, which gives us the probability of the z-score being less than -1.6. The probability is 0.0548 (approximately).