Check My Work (g remaining) eBook According to the National Association of Colle
ID: 3340353 • Letter: C
Question
Check My Work (g remaining) eBook According to the National Association of Colleges and Employers, the average starting salary for new college graduates in health sdiences is $51,541. The average starting salary for new colleg- raduates in business is $53,901 (National Association of Colleges and Employers website, January 5, 2015). Assume that starting salaries are normally distributed and that the standard deviation for starting salaries for new college graduates in health sciences is $11,000. Assume that the standard deviation for starting salaries for new college graduates in business is $15,000 Use Table 1 in Appendix B. a. What is That's Incorrect ollege graduate in business will earn a starting salary of at least $65,000 (to 4 decimals)? (Round z value to 2 decimal places.) b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000 (to 4 decimals)? (Round z value to 2 decimal places.) c. What is the probuability that a new college graduate in health sciences will carn a starting salary less than $40,000 (to 4 decimals)? (Round z value to 2 decimal places.) d How much would a new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences the nearest whole number)? Hide Feedhack Incorreet ere to searchExplanation / Answer
a.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 53901
standard Deviation ( sd )= 15000
P(X < 65000) = (65000-53901)/15000
= 11099/15000= 0.7399
= P ( Z <0.7399) From Standard Normal Table
= 0.7703
P(X > = 65000) = (1 - P(X < 65000)
= 1 - 0.7703 = 0.2297
b.
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 51541
standard Deviation ( sd )= 11000
P(X < 65000) = (65000-51541)/11000
= 13459/11000= 1.2235
= P ( Z <1.2235) From Standard Normal Table
= 0.8894
P(X > = 65000) = (1 - P(X < 65000)
= 1 - 0.8894 = 0.1106
c.
P(X < 40000) = (40000-51541)/11000
= -11541/11000= -1.0492
= P ( Z <-1.0492) From Standard Normal Table
= 0.147
d.
1% of new graduates in the health science earn minimum of
P ( Z > x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is 2.3263
P( x-u / (s.d) > x - 51541/11000) = 0.01
That is, ( x - 51541/11000) = 2.3263
--> x = 2.3263 * 11000+51541 = 77130.8266
& now we required to find a minimum of salary of 77130.8266 for new
graduate in business is -
P(X > 77130.8266) = (77130.8266-53901)/15000
= 23229.8266/15000 = 1.5487
= P ( Z >1.5487) From Standard Normal Table
= 0.0607