Consider the approximately normal population of heights of m

Consider the approximately normal population of heights of male college students with mean = 70 inches and standard deviation of = 3.5 inches. A random sample of 15 heights is obtained.

(a) Describe the distribution of x, height of male college students.

(b) Find the proportion of male college students whose height is greater than 69 inches. (Give your answer correct to four decimal places.)

(c) Describe the distribution of x, the mean of samples of size 15.

(d) Find the mean of the x distribution. (Give your answer correct to the nearest whole number.)


(ii) Find the standard error of the x distribution. (Give your answer correct to two decimal places.)


(e) Find P(x > 68). (Give your answer correct to four decimal places.)


(f) Find P(x < 70). (Give your answer correct to four decimal places.)

Solution

Consider the approximately normal population of heights of male college students with

mean = 70 inches and

standard deviation of = 3.5 inches.

A random sample of 15 heights is obtained.

n = 15

(a) Describe the distribution of x, height of male college students.

Let X be the height of male college students.

X has approximately Normal distribution with mean = 70 and = 3.5

(b) Find the proportion of male college students whose height is greater than 69 inches.

P(X > 69)

First we have to find z-score for x = 69,

z-score = (X - mean) /

z-score = (69 - 70) / 3.5 = -0.2857

Now we have to find P(Z > -0.2857).

P(Z > -0.2857) = 1 - P(Z <=-0.2857)

Because we can\'t find right tail probability directlyin EXCEL.

syntax :

=NORMSDIST(z)

where z is test statistic value.

P(Z <=-0.2857) = 0.3875

P(Z > -0.2857) = 1 - 0.3875 = 0.6125

(c) Describe the distribution of x, the mean of samples of size 15.

Let x be the mean of samples of size 15.

Let we denote it by Xbar.

Then Xbar has also Normal distribution with mean = 70 and

sd = SE = / sqrt(n)

sd = SE = 3.5/sqrt(15) = 0.90

where SE is the standard error.

(d) Find the mean of the x distribution.

mean of x is 70.

(ii) Find the standard error of the x distribution.

standard error (SE) = 0.90

(e) Find P(x > 68).

find z-score for x=68,

z = (68 - 70) / 3.5 = -0.5714

P(Z > -0.5714) = 1 - P(Z <= -0.5714)

P(Z <= -0.5714) = 0.2839

P(Z > -0.5714) = 1 - 0.2839 = 0.7161

(f) Find P(x < 70).

z-score for x = 70,

z = (70 - 70) / 3.5 = 0

Now we have to find P(Z < 0). This probability we can find directly in EXCEL.

syntax :

NORMSDIST(z)

where z is the test statistic value.

P(Z < 0) = 0.5000