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Math Help - probability/mean help please

  1. #1
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    probability/mean help please

    Hi every one i'm new here and i would really really appreciate it if you could help me with this problem:

    The vice-president of human resources of your company has created a new program of payment to reduce the amount of absenteeism. His strategy is to offer special extra payments at the end of each year to the employees who have not missed any day of work during the year.

    The company will give 1000$ to an employee at the end of one year without any missed day of work. this amount will be of 1500$ if it's the second year in a row without absenteeism, 2000$ if it's the third year in a row, 2500$ if it's the fourth year in a row and 5000$ if it's the fifth year in row. If an employee misses at least one day of work, he will be excluded of the program for the rest of the 5 years of the program.

    this new program will only last for five years, there are 1000 employees in this company and assume that there will be no firing or hiring while this program is applied.

    the probability to miss at least one day of work during
    ...the first year(of the program) is 0.10
    ...the first two years is 0.15
    ...the first three years is 0.35
    ...the first four years is 0.65
    ...the first five years is 0.85

    A)What is the total amount of money the company is expecting to pay for this program (the sum of the five years)

    That's my solution but i don't know if it's good or not:

    1000(1-0.10)(1000$)+ 900(1-0.15)(1500)+765(1-0.35)(2000)+ 497,25(1-1-0.65)+ 74.5875(1-0.85)(5000)

    B)What is the total amount of money the company is expecting to pay IF they allow the employees to participate every year of the program (for example: someone succeeds the first year (so earns 1000$) and the second year(and earns 1500$) but misses a day the third year (so earns 0$). this person would be allowed to participate the fourth year and fifth year but will be earning 1000$(for the fourth and 1500$ for the fifth if he succeeds) as if he was at the begging of the program)

    on this one i'm really not sure what to do, it's seems to be really long to solve but i'm not really sure how to approach this and would really really appreciate some help

    Thank you very much for reading all this and i hope i'll ''read'' from you soon

    sue

    P.S. i'm sorry if i'm a bit confusing to read sometimes but i have to translate from French to English sorry
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  2. #2
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    Wethersfield, CT
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    Hi:

    To me (and I may be off), the probability table says that 10% of the work force (1000) will miss >= one day in first yr. Similarly, 15% of workforce miss in years 1 and/or 2, 35% (1000) in yrs 1 and/or 2 and/or 3, etc.

    Since 100 fail in yr. 1, and 150 fail by the end of yr 2, then 50 fail in yr.2.
    350 failing by end of yr 3, and 150 previously eliminated, 200 fail during yr 3.
    Following this pattern, 300 and 200 are eliminated during yrs 4 and 5 respectively. Thus, bonus payouts saved by company are,

    yr 1: 100($1000) = $100k
    yr 2: 50($1500) = $75k
    yr 3: 200($2000) = $400k
    yr 4: 300($2500) = $750k
    yr 5: 200($5000) = $1 million
    ----------------------------
    $2,325,000 (total savings)

    Potential payout: 1000 win each year --> 1000[1000 + 1500 +...+ 5000]
    = $12,000,000.

    Expected payout: $12,000,000 - 2,325,000 = $9,675,000

    What do you think?

    Regards,

    Rich B.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Rich B. View Post
    Hi:

    To me (and I may be off), the probability table says that 10% of the work force (1000) will miss >= one day in first yr. Similarly, 15% of workforce miss in years 1 and/or 2, 35% (1000) in yrs 1 and/or 2 and/or 3, etc.
    My understanding is:

    90% of employees collect at the end of year 1
    85% of employees collect at the end of year 2
    65% of employees collect at the end of year 3
    35% of employees collect at the end of year 4
    15% of employees collect at the end of year 5

    RonL
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  4. #4
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    Hi,

    thanks for reading,

    A)For my part, i interpreted the different probabilities the same way CaptainBlack did:

    Quote Originally Posted by CaptainBlack View Post
    My understanding is:

    90% of employees collect at the end of year 1
    85% of employees collect at the end of year 2
    65% of employees collect at the end of year 3
    35% of employees collect at the end of year 4
    15% of employees collect at the end of year 5

    RonL

    So if 90% of the 1000 employees get the bonus at the end of the first year it means:

    (1000 employees)x(90%)x(bonus=1000$)

    for the second year, 85% of the 900 employees left from the first year would earn the bonus so:

    (85% of the 900 employees left= 0.85x900)x(bonus of the second year=1500$)

    and so on for the other years

    is my reasoning good? if i'm wrong could you please tell me which part exactely is

    thank you

    Sue
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by sue19 View Post
    Hi,

    thanks for reading,

    A)For my part, i interpreted the different probabilities the same way CaptainBlack did:




    So if 90% of the 1000 employees get the bonus at the end of the first year it means:

    (1000 employees)x(90%)x(bonus=1000$)

    for the second year, 85% of the 900 employees left from the first year would earn the bonus so:

    (85% of the 900 employees left= 0.85x900)x(bonus of the second year=1500$)

    and so on for the other years

    is my reasoning good? if i'm wrong could you please tell me which part exactely is

    thank you

    Sue
    Looks OK to me, but remember 90% is 0.90 when multiplying so 90% of 1000 employees
    receiving $1000 is $1000x0.90x1000=$900000

    RonL
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  6. #6
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    Thanks a lot for checking,

    But, do you have any idea for part B) of the problem?

    Sue
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