Question

Suppose the average height of all humans is normally distributed with a mean of 72 inches and a variance of 18 inches.

Answer 1

Answer:

a) P(75 < x < 80 ) = 0.2088

b) The probability that  average height of all humans less than 65

P( X < 65 ) = 0.0495

Step-by-step explanation:

Step(i):-

Given mean of the Population = 72

Given variance of the Population = 18 inches.

Standard deviation of the Population = √18 = 4.242

Let 'x' be the random variable in Normal distribution

a)

Given X₁ = 75

$Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{75-72}{4.242} = 0.70721$

Given X₂= 80

$Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{80-72}{4.242} = 1.885$

The probability that  average height of all humans between 75 and 80

$P(75 < X < 80 ) = P(0.70721 < Z < 1.885)$

                        = | A ( 1.885) - A( 0.70721|

                       = 0.4699 - 0.2611

                      = 0.2088

P(75 < x < 80 ) = 0.2088

b)

Step(ii):-

Given X₁ = 65

$Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{65-72}{4.242} = -1.650$

The probability that  average height of all humans less than 65

P( X < 65 ) = P( Z < - 1.650 )

                 = 1 - P( Z > 1.650)

                = 1 - ( 0.5 + A (1.650))

              =  0.5 - A( 1.65)

             = 0.5 - 0.4505

             = 0.0495

The probability that  average height of all humans less than 65

P( X < 65 ) = 0.0495