Question

The U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542. Suppose annual salaries in the metropolitan Boston area are normally distributed with a standard deviation of $4,246. A Boston worker is randomly selected. (Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

(a) What is the probability that the worker’s annual salary is more than $60,000?

(b) What is the probability that the worker’s annual salary is less than $42,000?

(c) What is the probability that the worker’s annual salary is more than $39,000?

(d) What is the probability that the worker’s annual salary is between $43,000 and $51,000?

Answer 1

Solution :

Given that,

mean = = 50,542

standard deviation = =4,246

a ) P (x > 60,000 )

= 1 - P (x < 60,000 )

= 1 - P ( x -  / ) < (60,000 - 50,542 / 4,246)

= 1 - P ( z < 9458 / 4,246 )

= 1 - P ( z < 2.23 )

Using z table

= 1 - 0.9938

= 0.0062

Probability = 0.0062

b ) P( x < 42,000 )

P ( x - / ) < ( 42,000 - 50,542 / 4,246)

P ( z <-8542 / 4,246 )

P ( z < -2.01)

= 0.0228

Probability = 0.0228

c ) P (x > 39,000 )

= 1 - P (x < 39,000 )

= 1 - P ( x -  / ) < (39,000 - 50,542 / 4,246)

= 1 - P ( z < -11542 / 4,246 )

= 1 - P ( z < - 2.72 )

Using z table

= 1 - 0.0030

= 0.9970

Probability = 0.9970

d ) P ( 43,000 < x < 51,000 )

P(43,000 - 50,542 / 4,246) < ( x -  / ) < (51,000 - 50,542 / 4,246)

P ( -7542 / 4,246 < z < 458 / 4,246 )

P (-1.78 < z < 0.11 )

P (z < 0.11 ) - P (z < -1.78 )

Using z table

= 0.5438 - 0.0375

= 0.5063

Probability = 0.5063