Question

The distribution of total body protein in adult men with liver cirrhosis is approximately normal with mean of 9.8 kg and a standard deviation of 0.1 kg.

(a) Find the probability that a randomly selected man with liver cirrhosis has a total body protein of 9.75 kg or greater?

(b) Find the probability that the mean of a group of 30 randomly selected men with liver cirrhosis has a total body protein of 9.75 kg or greater?

(c) What is the minimum total body protein that the upper 25% of adult men with cirrhosis have?

Answer 1

Solution :

Given that ,

mean = = 9.8

standard deviation = = 0.1

P( x > 9.75 ) = 1 - P( x < 9.75 )

=1- P[(x - ) / < ( 9.75 - 9.8 ) / 0.1 ]

=1- P( z < -0.5 )

Using z table,

= 1 - 0.3085

= 0.6915

Probability = 0.6915

( b )

n = 30

= 9.8

= / n = 0.1 / 30 = 0.0183

P( > 9.75 ) = 1 - P( < 9.75 )

= 1 - P[( - ) / < ( 9.75 - 9.8 ) / 0.0183 ]

= 1 - P( z < -2.73 )

Using z table,    

= 1 - 0.0032

= 0.9968

Probability = 0.9968

( c )

The z - distribution of the  25% is

P(Z > z) = 25%

=1- P(Z < z ) = 0.25

= P(Z < z ) = 1 - 0.25

P ( Z < z ) = 0.75

P ( Z < 0.674 ) = 0.75

z = 0.674

Using z-score formula,

x = z * +

x = 0.674 * 0.1 + 9.8

x = 9.8674

x = 9.87