Question

II. In an impoverished urban area, individual income is normally distributed with a mean of $21,000 and a standard deviation of $2,000. What percentage of incomes fall: a. Between $19,000 and $26,000?

b. Between $23,000 and $26,000?

c. Between $16,000 and $18,000 ?

d. Below $16,000?

e. Between $15,000 and $27,000?

f. What is the probability of locating someone with an income below $16,000 per year?

Answer 1

Solution:

Given that,

   = 21000

= 2000

a ) p (19000 < x  < 26000 )

= p( 19000 - 21000 / 2000 ) ( x -  / ) < ( 26000 - 21000 / 2000)

= p ( -2000 /2000 < z < 5000 /2000 )

= p ( - 0 < z < 2.5)

= p (z < 2.5 ) - p ( z < - 0 )

Using z table

= 0.9938 - 0.5000

= 0.4938

Probability = 0.4938

b ) p (23000 < x  < 26000 )

= p( 23000 - 21000 / 2000 ) ( x -  / ) < ( 26000 - 21000 / 2000)

= p ( 2000 /2000 < z < 5000 /2000 )

= p ( 0 < z < 2.5)

= p (z < 2.5 ) - p ( z < 0 )

Using z table

= 0.9938 - 0.5000

= 0.4938

Probability = 0.4938

d ) p ( x  < 16000 )

= p ( x -  / ) < ( 16000 - 21000 / 2000)

= p ( z < - 5000 /2000 )

= p ( z < - 2.5)

Using z table

= 0.0062

Probability = 0.0062

e ) p (15000 < x  < 27000 )

= p( 15000 - 21000 / 2000 ) ( x -  / ) < ( 27000 - 21000 / 2000)

= p ( - 6000 /2000 < z < 6000 /2000 )

= p ( - 3 < z < 3 )

= p (z < 3 ) - p ( z < - 3 )

Using z table

= 0.9987 - 0.0013

= 0.9974

Probability = 0.9974

f ) p ( x  < 16000 )

= p ( x -  / ) < ( 16000 - 21000 / 2000)

= p ( z < - 5000 /2000 )

= p ( z < - 2.5)

Using z table

= 0.0062

Probability = 0.0062