Question

1.     The distribution of systolic blood pressure among college students has a mean of 114.6 and a standard deviation of 14.1. Assuming the systolic blood pressure among college students is normally distributed. 

What is the probability that someone selected at random:

a)        Has a blood pressure greater than 120?

b)        Has a blood pressure less than 110?

c)        Has a blood pressure between 105 and 125?

d)        Has a blood pressure outside of the interval between 100 and 130?

e)        95% of the students have their blood pressures above what value?

f)         Between what two symmetrically distributed blood pressures around the mean do 90% of the systolic blood pressures occur?

g)        If a recording of blood pressure had only one reading of 100, what is the chance that it was a systolic blood pressure recording (for it might be the recording of a diastolic blood pressure as well)?

Answer 1

Let X denote the systolic pressure in college students, which is normally distributed, so

X\simN( 114.6, 14.12 )

(a) P[X>120] = P[(X-114.6)/14.1 > (120-114.6)/14.1 ] = P[Z > .38297] = 1-P[Z<.38297] = 1-.64803 = 0.35197

(b) P[X<110] = P[(X-114.6)/14.1<(110-114.6)/14.1] = P[Z< -0.3262) = P[Z>0.3262] = 1-P[Z<0.3262] = 1-0.62930 = 0.3707

(c) P[105<X<125] = P[(105-114.6)/14.1 < (X-114.6)/14.1 < (125-114.6)/14.1]

= P[-0.6809<Z<0.7375]

= P[Z<0.7375]- P[Z<-0.6809]

= P[Z<0.7375] - P[Z>0.6809]

= P[Z<0.7375] - {1-P[Z<0.6809]}

= P[Z<0.7375] - 1 + P[Z<0.6809]

= 0.77035 - 1 + 0.75175

= 0.5221

(d) 1 - P[100<X<130] = 1 - P[-1.035<Z<1.0921]

= 1 - { P[Z<1.0921] - P[Z<-1.035] }

= 1 - { P[Z<1.0921] - P[Z>1.035] }

= 1- { P[Z<1.0921] - {1-P[Z<1.035] } }

= 1-{ P[Z<1.0921] - 1 + P[Z<1.035] }

= 1-P[Z<1.0921] +1 - P[Z<1.035]

= 1-0.86214+1-0.85083

= 0.28703

(e) let the value be M

= P[X > M] = 0.95

= P[Z > (M-114.6)/14.1] = 0.95

= 1- P[Z< (M-114.6)/14.1] = 0.95

= P[Z< (M-114.6)/14.1] = 0.05 (1)

now we know that P[Z<1.6449] = 0.05 (from normal tables) (2)

On Comparing (1) and (2)

we get

= (M-114.6)/14.1 = 1.6449

= M = 137.8

(g) prob of that recording to be systolic is 1/2.