Question

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.)

Answer #1

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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