Question

On an ostrich farm, the weights of the birds are found to be normally distributed. The weights of the females have a mean 78.6 kg and a standard deviation of 5.03 kg. The weights of the males have a mean 91.3 kg and a standard deviation of 6.29 kg. Find the probability that a randomly selected: Male will weight less than 80 kg. Female will weight less than 80 kg. Female will weigh between 70 and 80 kg. 20% of females weigh less than k kg. Find k. The middle 90% of the males weigh between a kg and b kg. Find the values of a and b kg.

Answer 1

Solution :

Given that ,

male

1) mean = = 91.3

standard deviation = = 6.29

P(x < 80 ) = P[(x - ) / < (80- 91.3) /6.29 ]

= P(z < -1.80)

= 0.0359

probability =0.0359

2) female

mean = = 78.6

standard deviation = = 5.03

a)

P(x < 80 ) = P[(x - ) / < ( 80-78.6 )  / 5.03 ]

= P(z < 0.28)

= 0.6103

probability =0.6103

b)

P( 70< x <80 ) = P[(70-78.6)/5.03 ) < (x - ) /  < (80 -78.6) / 5.03) ]

= P( -1.71< z < 0.28 )

= P(z < 0.28 ) - P(z < -1.71 )

Using standard normal table

= 0.6103 - 0.0436 = 0.5666

Probability = 0.5666

c)

P(Z <k) = 0.20

P(Z < -0.84) = 0.20

k= -0.84

middle 90%

P(-z Z z) = 0.90

P(Z z) - P(Z -z) = 0.90

2P(Z z) - 1 = 0.90

2P(Z z) = 1 + 0.90 = 1.90

P(Z z) = 1.90 / 2 = 0.95

P(Z 1.645 ) =0.95

z = - 1.645 , +1.645

z = - 1.645

Using z-score formula,

x = z * +

x =-1.645*6.29+91.3

x = 80.95

z = 1.645

x = z * +

x =1.645*6.29+91.3

x = 101.65

Value a and b = 80.95 and 101.65