Among females in the United States between 18 and 74 years of age, the diastolic

Question

Among females in the United States between 18 and 74 years of age, the diastolic blood pressure is normally distributed with the = 77 mmHg and = 11.6 mmHg. a) For a randomly selected woman from this population, what is the probability that she will have a diastolic blood pressure above 90mmHg? b) What is the probability that she will have a diastolic blood pressure between 60 and 90 mmHg?
Please show all work. Thanks! Among females in the United States between 18 and 74 years of age, the diastolic blood pressure is normally distributed with the = 77 mmHg and = 11.6 mmHg. a) For a randomly selected woman from this population, what is the probability that she will have a diastolic blood pressure above 90mmHg? b) What is the probability that she will have a diastolic blood pressure between 60 and 90 mmHg?
Please show all work. Thanks! a) For a randomly selected woman from this population, what is the probability that she will have a diastolic blood pressure above 90mmHg? b) What is the probability that she will have a diastolic blood pressure between 60 and 90 mmHg?
Please show all work. Thanks!

Explanation / Answer

for normal distribution z=(X-mean)/std deviation

hence with help of normal value table:

a)probability that she will have a diastolic blood pressure above 90mmHg=P(X>90)=1-P(X<90)=1-P(Z<(90-77)/11.6)

=1-P(Z<1.1207)=1-0.8688 =0.1312

b) P(60<X<90)=P((60-77)/11.6<Z<(90-77)/11.6)=P(-1.4655<Z<1.1207)=0.8688-0.0714 =0.7974

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