Suppose that diastolic blood pressure in hypersensitive women centers at 100mmHg and has a standard deviation of 16mmHg and it is normally distributed. What is the probability of randomly selecting a woman with a diastolic blood pressure less than 92mmHg? Provide your answer to four decimal places.
(92 -100)/16 = z-score/
-8/16 =-.5
Now you either have to use a z-table to find the values below -.5
I used a Ti-84 calculator and got .3085. See if you can get this answer.
To solve this problem, we need to find the probability of randomly selecting a woman with a diastolic blood pressure less than 92mmHg.
First, we need to standardize the value of 92mmHg using the formula for standardization:
Z = (X - μ) / σ
Where:
Z is the standard score
X is the diastolic blood pressure value (92mmHg in this case)
μ is the mean of the diastolic blood pressure distribution (100mmHg)
σ is the standard deviation of the diastolic blood pressure distribution (16mmHg)
Now we substitute the values into the formula:
Z = (92 - 100) / 16
Calculating this, we get:
Z = -0.5
Now, we can use the standardized Z-score to find the probability using a standard normal distribution table or a calculator.
Using either of these tools, we can find that the probability corresponding to a Z-score of -0.5 is approximately 0.3085.
Therefore, the probability of randomly selecting a woman with a diastolic blood pressure less than 92mmHg is 0.3085 when rounded to four decimal places.