Consider the approximately normal population of heights of male college students with mean μ = 72 inches and standard deviation of σ = 3.2 inches. A random sample of 21 heights is obtained.
(a) Describe the distribution of x, height of male college students.
skewed rightapproximately normal chi-squareskewed left
(b) Find the proportion of male college students whose height is
greater than 74 inches. (Round your answer to four decimal
places.)
(c) Describe the distribution of x, the mean of samples of
size 21.
skewed leftskewed right chi-squareapproximately normal
(d) Find the mean of the x distribution. (Round your
answer to the nearest whole number.)
(ii) Find the standard error of the x distribution. (Round
your answer to two decimal places.)
(e) Find P(x > 71). (Round your answer to four
decimal places.)
(f) Find P(x < 68). (Round your answer to four
decimal places.)
µ = 72, σ = 3.2, n = 21
(a) Distribution of x: approximately normal
(b) Proportion of male college students whose height is greater than 74 inches, P(X > 74) =
= P( (X-µ)/σ > (74-72)/3.2)
= P(z > 0.625)
= 1 - P(z < 0.625)
Using excel function:
= 1 - NORM.S.DIST(0.625, 1)
= 0.266
(c) Distribution of x, the mean of samples of size 21: approximately normal
(d) Mean of the x distribution, µₓ = 72
(ii) Standard error of the x distribution, σₓ = σ/√n = 3.2/√21 = 0.6983 = 0.70
(e) P(X > 71) =
= P( (X-µ)/σ > (71-72)/3.2)
= P(z > -0.3125)
= 1 - P(z < -0.3125)
Using excel function:
= 1 - NORM.S.DIST(-0.3125, 1)
= 0.6227
(f) P(X < 68) =
= P( (X-µ)/σ < (68-72)/3.2 )
= P(z < -1.25)
Using excel function:
= NORM.S.DIST(-1.25, 1)
= 0.1056
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