Question

Systolic Blood Pressure Assume that the mean systolic blood pressure of normal adults is 120 millimeters of mercury and the standard deviation is 5.6. Assume the variable is normally distributed. Use a TI-83 Plus/TI-84 Plus calculator and round the answers to at least four decimal places.
If an individual is selected, find the probability that the individual's pressure will be between 119.4 and 121.4mmHg
P (119.4 50) =?

Answer 1

Answer:

0.1425 = 14.25% probability that the individual's pressure will be between 119.4 and 121.4mmHg.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 120, \sigma = 5.6$

Find the probability that the individual's pressure will be between 119.4 and 121.4mmHg

This is the pvalue of Z when X = 121.4 subtracted by the pvalue of Z when X = 119.4. So

X = 121.4

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{121.4 - 120}{5.6}$

$Z = 0.25$

$Z = 0.25$ has a pvalue of 0.5987

X = 119.4

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{119.4 - 120}{5.6}$

$Z = -0.11$

$Z = -0.11$ has a pvalue of 0.4562

0.5987 - 0.4562 = 0.1425

0.1425 = 14.25% probability that the individual's pressure will be between 119.4 and 121.4mmHg.