Answer to Question #180930 in Statistics and Probability for Richard Sasunathan

Question #180930

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 = πŸπŸ”πŸ mm Hg and standard deviation, 𝝈 = πŸπŸ“ mm Hg. Assume the systolic blood pressure is normally distributed.

a. What is the probability of patients with a systolic blood pressure of more than 180 mm Hg?

b. How many patients will have a systolic blood pressure of more than 180 mm Hg?

c. What is the probability of patients with a systolic blood pressure between 145 and 160 mm Hg?

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

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

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


1
Expert's answer
2021-04-15T06:50:58-0400

Let X denotes the systolic blood pressure. The X follows normal with "\\mu = 161, \\sigma = 25"

Let "Z = \\frac{X - \\mu}{\\sigma}" ~N(0,1)

a. P(X>180)

"P(X>x) = P(\\frac{X-\\mu}{\\sigma}>\\frac{x-\\mu}{\\sigma}) = P(Z>\\frac{x-\\mu}{\\sigma}) = 1 - P(Z\\le \\frac{x-\\mu}{\\sigma})"

Therefore, when x = 180,

"\\frac{x-\\mu}{\\sigma}=\\frac{180-161}{25}=0.76 \\\\\n\n\\therefore P(X>180) = 1 - P(Z\\le 0.76) = 1 - 0.7764 = 0.2236"

b. Number of patients will have systolic blood pressure more than "180 = 0.2236 \\times 10000 = 2236"

c. "P(145 \\le X \\le 160)"

Now,

"P(x_1\\le X \\le x_2) = P(\\frac{x_1-\\mu}{\\sigma}\\le \\frac{X-\\mu}{\\sigma} \\le \\frac{x_2-\\mu}{\\sigma}) \\\\\n\nP(x_1\\le X \\le x_2) = P(\\frac{x_1-\\mu}{\\sigma}\\le Z \\le \\frac{x_2-\\mu}{\\sigma}) \\\\\n\nP(x_1\\le X \\le x_2) = P(Z \\le \\frac{x_2-\\mu}{\\sigma}) - P(Z \\le \\frac{x_1-\\mu}{\\sigma})"

when x = 145,

"\\frac{x-\\mu}{\\sigma}=\\frac{145-161}{25}=-0.64"

when x = 160,

"\\frac{x-\\mu}{\\sigma}=\\frac{160-161}{25}=-0.04"

"P(145 \\le X \\le 160) = P(Z\\le-0.04) - P(Z\\le -0.64) = 0.2229"

d.

i. The sampling distribution for mean with sample size "= 60 \\times 30 = 1800" is

"N(161,\\sqrt{1800}) = N(161,42.4264)"

ii. "P(X\\le 165)"

"P(X\\le x) = P(\\frac{X-\\mu}{\\sigma}\\le \\frac{x-\\mu}{\\sigma}) = P(Z\\le \\frac{x-\\mu}{\\sigma})"

When x = 165

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


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