Consider the approximately normal population of heights of m

Consider the approximately normal population of heights of male college students with mean = 68 inches and standard deviation of = 5 inches. A random sample of 12 heights is obtained.

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

skewed right approximately normal     skewed left chi-square

(b) Find the proportion of male college students whose height is greater than 69 inches. (Give your answer correct to four decimal places.)
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(c) Describe the distribution of x, the mean of samples of size 12.

skewed right approximately normal     skewed left chi-square

(d) Find the mean of the x distribution. (Give your answer correct to the nearest whole number.)
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(ii) Find the standard error of the x distribution. (Give your answer correct to two decimal places.)
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(e) Find P(x > 69). (Give your answer correct to four decimal places.)
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(f) Find P(x < 65). (Give your answer correct to four decimal places.)
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Solution

We have given that the approximately normal population of heights of male college students with mean = 68 inches and standard deviation of = 5 inches.

A random sample of 12 heights is obtained.

Describe the distribution of x, height of male college students.

x is height of male college students.

x has also approximately normal distribution.

Because sample is the subset of population then whatever be the distribution of the population that will be the distribution of sample.

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

Given that X is height of the male college students.

P(X > 69) = P ( (X- mean)/sd > (69 - mean) /sd )

= P (Z > ( 69 - 68 ) / 5 )

= P(Z > 0.2) = 1 - P (Z <= 0.2) = 1 - 0.5793 = 0.4207

Describe the distribution of x, the mean of samples of size 12.

We know that by Central limit theorem the distribution of the mean of samples is Approximately normal with mean is the mean of original population (µ) and standard deviation is /sqrt(n).

where µ = 68

standard deviation = 5/ sqrt(12) = 1.4434

The mean of the x distribution is 68

  The standard error (SE)of the x distribution is 1.44

Find P(x > 69).

where x is the distribution of the mean of samples.

P(x > 69) = P( (x - mean) / SE > (69 - mean) / SE )

= P(Z > ( 69 - 68 ) / 1.44 )

= P( Z > 0.69) = 1 - P( Z <=0.69) = 1 - 0.7549 = 0.2451

Find P(x < 65). (Give your answer correct to four decimal places.)

P(x < 65) = P( (x - mean) / SE < (65 - mean) / SE )

= P( Z < (65 - 68) / 1.44)

= P( Z < - 2.08) = 0.0188

The probabilities we can find by using EXCEL :

syntax : =normsdist(z)