Question

Assume that the mean systolic blood pressure of normal adults is 120 millimeters of mercury (mm Hg) and the population standard deviation is 5.6. Assume the variable is normally distributed. If a sample of 30 adults is randomly selected, find the probability that the sample mean will be between 120-mm and 121.8-mm Hg.

Answer #1

**Solution:**

Given that

mean μ = 120 standard deviation σ = 5.6

P(120 < X < 121.8) = P(X < 121.8) - P(X < 120)

= P(Z < (121.8-120)/5.6) - P(Z < (120-120)/5.6)

= P(Z < 0.32) - P(Z < 0)

= 0.625516 - 0.500000 = 0.125516

mean μ = 120 standard deviation σ = 5.6, n = 30

P(120 < X < 121.8) = P(X < 121.8) - P(X < 120)

= P(Z < (121.8-120)/(5.6/sqrt(30)) - P(Z <
(120-120)/(5.6/sqrt(30))

= P(Z < 1.76) - P(Z < 0)

= 0.960796 - 0.500000 = 0.460796

The mean systolic blood pressure of adults is 120 millimeters of
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2) If a sample of 30 adults are randomly selected, what is the
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-Central Limit...

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