Question

2) 030 points defined as a eystolic Hond pressure shove 14 a) What io the prehability a rendomty selected wman betwn the nges of 1s-24 will be elassified as being hypertensbe b) In a random sample of 4 women (betwecen ages 18-24)what is the probabilitys the sample mean is r than 1407 c) What is the Interquartile Range (IQR) for the systolic blood pressures of women between 18- 24? d) In a random sample of 4 women between the ages of 18-24, what is the expected u.ss value and standard error of the sample mean? e) What is the probability a randomly selected woman between the ages of 18-24 will have a systolic blood pressure greater than Q+1.5 IQR?

Answer 1

a) P(X > 140)

= P((X - $\mu$)/$\sigma$ > (140 - $\mu$)/$\sigma$)

= P(Z > (140 - 114.8)/13.1)

= P(Z > 1.92)

= 1 - P(Z < 1.92)

= 1 - 0.9726

= 0.0274

b) P($\bar x$ > 140)

= P(($\bar x$ - $\mu$)/($\sigma/\sqrt n$) > (140 - $\mu$)/($\sigma/\sqrt n$))

= P(Z > (140 - 114.8)/(13.1/))

V4

= P(Z > 3.85)

= 1 - P(Z < 3.85)

= 1 - 1 = 0

c) P(X < x) = 0.25

or, P((X - $\mu$)/$\sigma$ < (x - $\mu$)/$\sigma$) = 0.25

or, P(Z < (x - 114.8)/13.1) = 0.25

or, (x - 114.8)/13.1 = -0.67

or, x = -0.67 * 13.1 + 114.8

or, x = 106.023

P(X < x) = 0.75

or, P((X - $\mu$)/$\sigma$ < (x - $\mu$)/$\sigma$) = 0.75

or, P(Z < (x - 114.8)/13.1) = 0.75

or, (x - 114.8)/13.1 = 0.67

or, x = 0.67 * 13.1 + 114.8

or, x = 123.577

IQR = Q3 - Q1 = 123.577 - 106.023 = 17.554

d) $\mu_{\bar x}$ = 114.8

   $\sigma_{\bar x}$ = $\sigma/\sqrt n$

       = 13.1/ = 6.55

V4

e) Q3 + 1.5 * IQR = 123.577 + 1.5 * 17.554 = 149.908

P(X > 149.908)

= P((X - $\mu$)/$\sigma$ > (149.908 - $\mu$)/$\sigma$)

= P(Z > (149.908 - 114.8)/13.1)

= P(Z > 2.68)

= 1 - P(Z < 2.68)

= 1 - 0.9963

= 0.0037