Question

The salaries of the employees of a corporation are normally distributed with a mean of $25,000 and a standard deviation of $5,000.

a. What is the probability that a randomly selected employee will have a starting salary of at least $31,000?

b. What percentage of employees has salaries of less than $12,200?

c. What are the minimum and the maximum salaries of the middle 95% of the employees?

d. If sixty-eight of the employees have incomes of at least $35,600, how many individuals are employed in the corporation?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 25000

standard deviation = $\sigma$ = 5000

P(x $\geq$ 31000 ) = 1 - P(x  $\leq$ 31000 )

= 1 - P[(x - $\mu$ ) / $\sigma$ $\leq$ ( 31000 - 25000 ) / 5000 ]

= 1 -  P(z $\leq$ 1.2 )   

  Using z table,

= 1 - 0.8849

= 0.1151

Probability = 0.1151

( b)

P(x < 12200 ) = P[(x - $\mu$ ) / $\sigma$ < ( 12200 -25000 ) / 5000 ]

= P(z < -2.56 )

Using z table,

= 0.0052

= 0.52%

Answer = 0.52%

( c )

Middle 95%   

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 = 0.05

$\alpha$/2 = 0.025

1 - $\alpha$ /2 = 1 - 0.025 = 0.975

Z$\alpha$/2 = Z0.025 = -1.96

Z 1 - $\alpha$ /2 =Z0.975 = 1.96

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -1.96 * 5000 + 25000

x = 15200

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.96 * 5000 + 25000

x = 34800

Answer = minimum = 15200 and the maximum salaries = 34800

( d )

P(x $\geq$ 35600 ) = 1 - P(x  $\leq$ 35600)

= 1 - P[(x - $\mu$ ) / $\sigma$ $\leq$ ( 35600 - 25000) / 5000]

= 1 -  P(z $\leq$   2.12 )   

  Using z table,

= 1 - 0.9830

= 0.0170 * 68

= 1.156

Answer x = 1