The Diastolic Blood Pressure (DBP) of the USA middle-aged men population is norm

Question

The Diastolic Blood Pressure (DBP) of the USA middle-aged men population is normally distributed with mean µ = 80mm HG (millimeters of mercury as a measurement of pressure), and standard deviation = 12 (hence variance 2 = 144). (This number is the denominator in, for example, 120/80, where the numerator is the systolic pressure.)

(a) What proportion of such men have a DBP above 90?

(b) What proportion of such men have a DBP below 70?

(c) What proportion of such men have a DBP between 56 and 104?

(d) Find the value x for which the proportion of such men that have a DBP above x is equal to 0.25.

(e) For what value of x does it hold that the proportion of such men that have a DBP between 80 x and 80 + x is equal to 0.997?

Explanation / Answer

we need to make use of z tables to answer this , please keep them handy

Please use the formula as

Z = (X - Mean )/SD , where X = 80 and SD = 12

a) we need to find p(X>90)

Z = (90-80)/12 = 10/12 = 0.833

so we calclulate the value as

P ( Z>0.833 )=1P ( Z<0.833 )=10.7967=0.2033

b) again P(X<70)

Z = (70-80)/12 = -0.833

so P(Z<-0.833)

P ( Z<0.833 )=1P ( Z<0.833 )=10.7967=0.2033

c)

P(56<X<104)

so again Z = (56-80)/12 and (104-80)/12

so P( -2<z<2)

We see that P ( Z<2 )=0.9772.

P ( Z<2 ) can be found by using the following fomula.

P ( Z<a)=1P ( Z<a )

After substituting a=2 we have:

P ( Z<2)=1P ( Z<2 )

We see that P ( Z<2 )=0.9772 so,

P ( Z<2)=1P ( Z<2 )=10.9772=0.0228

At the end we have:

P (2<Z<2 )=0.9544

d) now this time poportion is given as 0.25 and we need to find x as

0.25 = (X - 80)/12

X - 0.25*12 + 80

= 83

We can answer only 4 subparts of a question at a time , as per the answering guidelines

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