The annual salary for a group of highlyeducated professional

The annual salary for a group of highly-educated professional is normally distributed with mean u=$138,000. Is also known that 62% earn more than $125,000. Compute answer in Excel/PHstat.

a. Determine the standard deviation.

b. What percent earn less than $165,000?

c. What percent earn between $125,000 and $145,000?

d. Determine the annual salary X1 for which 10% earn more than X1?

e. A random sample of 12 professionals is chosen from the group. What is the probability that at most 2 make more than $150,000?

f. A random sample of 25 professionals is chosen from the group. What is the probability that the mean salary of the sample is more than $155,000?

Solution

a) P [ X > 125000 ] = 0.62..
i.e, P [ Z > ( 125000 - 138000) / S.D ] = 0.62
So, -0.3054808 = -13000 / s.d

or, s.d = 42555.87.....
b) p [ x < 165000) = p [ z < ( 165000-138000) / 42555.87 ] = p [ z < 0.6344601 ] = 0.7371097 = 73.71%....
c( p [ 125000 < x < 145000 ] = p [ (125000-138000) / 42555.87 < z < ( 145000-138000) / 42555.87 ]
= p [ -0.3054808 < z <  0.1644896 ]

= p [ z < 0.1644896 ] - p [ z <   -0.3054808 ] =  0.1853271..i.e, 18.53%...

d) 90 percentile = 138

= 138000 + ( 1.281552 * 42555.87 ) =192537.6(ans)

e) P [ X > 150000 ] = P [ Z > ( 150000 - 138000) / (42555.87/ sqrt(12) ) ] = p [ z > 0.9768152 ] = 0.1643303

so, prob. of any1 person from 12 make more than 150000 is 0.1643303

probability that at most 2 make more than $150,000
= ( 12 C 0) ( 0.1643303)^0 (1 - 0.1643303)^12 + (12 C 1 ) (0.1643303 )^1 * ( 1- 0.1643303) ^ 11 + ( 12 C 2 ) ( 0.1643303)^2 * ( 1 - 0.1643303) ^ 10

=  0.6857153.........

F) P [ mean salaray > 155000 ] = p [ z > ( 155000 - 138000) / (42555.87/ sqrt(25) ) ]
= p [ z > 1.997374 ] =  0.02289229......