Can i get some help with these porblems? Can it be show in excel with the formul
ID: 3174981 • Letter: C
Question
Can i get some help with these porblems? Can it be show in excel with the formulas showing.
When people donate blood, their blood pressures are measured. Abnormal diastolic blood pressure (the lower of the two numbers) is related to strokes, heart attacks, and vascular disease. In the US, diastolic blood pressure is normally distributed such that the mean is 77 mmHg and the standard deviation is 12 mmHg. If a person has diastolic blood pressure (consistently) higher than 90 mmHg, then they are said to have hypertension (high blood pressure). Compute the probability that a randomly selected person in the US will have a diastolic blood pressure higher than 90 mmHg If a person has a diastolic blood pressure higher than 100 mmHg or lower than 50 mmHg, then they are not allowed to donate blood. Compute the probability that a randomly selected person in the US will have a diastolic blood pressure higher than 100 mmHg or lower than 50mmHg. If a group of 300 Americans are randomly seleced to donate blood, about how many would be expected to be disqualified because of diastolic blood pressure? A group of medical students believes that their college's students' diastolic blood pressures are different from the general population. They take a sample of 50 students from their college. If the students' blood pressures are not different from the general US population, what is the proability that the mean of a 50 student sample would be less than 75 mmHg?Explanation / Answer
a) given mean = 77 and sd = 12
so z score is
Z = (X - mean)/sd
now we need to find P(X>90)
so (90-77)/12 = 1.083 , from the z tables we get the probability as
P ( Z>1.083 )=1P ( Z<1.083 )=10.8599=0.1401
b) P( X>100) or P(X<50)
(100-77)/12 or (50-77)/12 solving this we get
P(Z>1.91) or P(Z<-2.25)
again from the z tables we get
P ( Z>1.91 )=1P ( Z<1.91 )=10.9719=0.0281
and P ( Z<2.25 )=1P ( Z<2.25 )=10.9878=0.0122 , since we are considering an OR here , lets add the probabilities together
0.0122+0.0281 =0.0403
c) if n = 300 , then based on the probability calculation above the number of people who will disquallify are
300*0.0403 = 12.09 , approximately 12
d) n = 50
again modifying the z score as
Z = (X-mean)/(sd/sqrt(n))
= (75-77)/(12/sqrt(50)) = -1.17
so P(Z<-1.17), again using the z tables we calculate the values as
will use the following formula:
P ( Z<a ) =1P ( Z<a )
If we put a=1.17 we get:
P ( Z<1.17 )=1P ( Z<1.17 )
Now we can find P ( Z<1.17 ) by using the standard normal table.
Since P ( Z<1.17 )=0.879, we conclude that:
P ( Z<1.17 )=1P ( Z<1.17 )=10.879=0.121