Answer to Question #183763 in Statistics and Probability for Richy

Question

Richard

Question 2

In a study of hypertension and optimal treatment conducted by the National Heart Institute, 10,000 patients had a mean systolic blood pressure (BP), 𝝁 = 𝟏𝟔𝟏 mm Hg and standard deviation, 𝝈 = 𝟐𝟓 mm Hg. Assume the systolic blood pressure is normally distributed.

d. If 60 random samples each of size 30 are drawn from this population, determine:

i. the sampling distribution of the mean systolic blood pressure.

ii. the probability that the of mean systolic blood pressure between 140 and 165 mm Hg.

Hi, regarding question d ( ii) based from#180930 / how you calculate the answer P(0.4478) and also the answer 0.5376, can show me the exact calculations. tq

When x = 165

\frac{x-\mu}{\sigma}=\frac{x-161}{42.4264}=0.4478 \\ \therefore P(Z\le 0.4478) = 0.5376

σ

x−μ

​

=

42.4264

x−161

​

=0.4478

∴P(Z≤0.4478)=0.5376

Answer 1

We have,

$\mu = 161$

$\sigma = 25$

Since, the systolic blood pressure is normally distributed

i) $f(x) = \dfrac{1}{\sigma \sqrt{2\pi}}e^-\dfrac{1}{2}(\dfrac{x- \mu}{\sigma})^2$

$f(x) = \dfrac{1}{25\sqrt{2\pi}}e^-\dfrac{1}{2}(\dfrac{x- 161}{25})^2$

ii) $P(140<X<165)$

$= P(\dfrac{140-165}{25}<Z< \dfrac{165-161}{25})$

$=P(-1<Z<0.16)$

$= P(-1<Z<0)+P(0<Z<0.16)\\= P(0<Z<1)+P(0<Z<0.16)$

$= 0.0398+0.636$

$= 0.67$