(a) (1 pt) Assume diastolic blood pressure is normally distributed in a certain population, with a mean of 82 mmHg and a standard deviation 11 mmHg. What proportion of this population has a diastolic blood pressure less than 60 mmHg (i. e., what is the probability that a person in this population has a diastolic blood pressure less than 60)?

2.275% of this population has a diastolic blood pressure less than 60 mmHg

Explanation:

Hello!

Yo have the distribution of the diastolic blood pressure in a certain population. Be X: diastolic blood pressure of an individual, X~N(μ;δ²)

Where

μ= 82mmHg

δ=11 mmHg

You need to calculate the probability of an individual of this population having less than 60mmHg diastolic blood pressure.

Symbolically:

P(X<60)

To obtain the value of probability you need to standardize the value of diastolic pressure so that you can obtain it from the standard normal distribution. The way to standardize the value is to subtract the mean and divide by the standard deviation

Z= (X-μ)/δ~N(0;1)

P(Z<(60-82)/11)

P(Z<-2)= 0.02275

I hope it helps!

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