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Math Help - Prob Homework help any of these 6 questions

  1. #1
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    Prob Homework help any of these 6 questions

    Hi this is my first time in the forums, and it'll be great to get some help.
    My teacher does not explain well even when questions are asked. It would be great to know how it is done.

    1. A bakery has 100 bags of cookies in stock. Each bag contains 14 chocolate chip cookies. The probability that a cookie have fewer than 10 chips in it is 10%.
    Therefore:
    100 bags
    14 cookies in each bag
    =
    1400 total cookies
    140 cookies with less than 10 chips

    -a)Find the probability in a bag that there are exactly 3 cookies that have fewer than 10 chips in them.

    -b)What is the expected number of cookies, that have fewer than 10 chips, in a bag?

    -c)What is the standard deviation of the number of cookies, that have fewer than 10 chips, in a bag?

    -d) How many of the 100 bags of cookies would you expect to have exactly 3 cookies that have fewer than 10 chips?

    -e) What is the probability that between 14 and 25 (inclusive) bags out of the 100 bags have exactly 3 cookies with fewer than 10 chips?

    ---------------------------------------------------
    2.
    -a)In a amusement park, youngsters jump around in a chamber full of balls. That sign in front of this attraction indicates that only those folks of weight between 45 lbs and 100 lbs are allowed in the chamber. Assume the weight of a youngster between 8 to 12 is normally distributed with mean 70lbs and standard deviation 10 lbs, what is the percentage of youngsters allowed in the chamber?

    -b)The youngsters are allowed in the chamber for 5 minutes per round, but the line of youngsters waiting to get in has a waiting time normally distributed with mean of waiting equal to 12 minutes and standard deviation of 4 minutes. Your son would love to jump around in the chamber but you have only 15 minutes to spare before the next shuttle bus comes in. What is the probability that your son could get in the pit of balls.

    --------------------------------------------------
    3. A Electro-mechanical Toothbrush (EMT) that claims to be 80% effective in fighting formation of plaque on teeth. GH consumer group that oversees that welfare of the consumer believe that the device is only 60% effective. A test is done on 10 people; if 7 or more people is free of plaque after using the device for 3 months, then GH will give EMT the seal of approval

    -a) What is the probability that EMT will be errorneously approved when in fact GH is correct in its belief?

    -b) What is the probability that EMT will be denied of approval when in fact its claim?

    ---------------------------------------------------
    4. Prostate cancer is a slow growing disease that strike 1 in 5 males at age between 50 and 60. A not too sensitive test(PSA) is available to test the existence of prostate sensitive antigens and is found to be accurate only 80% of the time when given to an afflicted person and also when given to a healthy person.

    -a) Draw a two-tier tree diagram and find the probability of the false positive (ie. the probability of being healthy and tested positive by the test). Also find the conditional probability of being healthy even when tested positive by the test)

    -b) James after tested once positive was tested again. Fill out a three-tier tree diagram and find the probability of false double-positive (ie. the probability of a person being healthy and twice-positively tested).
    Aso find the conditional probability of healthy even when John was tested positive twice.

    ---------------------------------------
    5. Two fair dice, each with seven faces are rolled. Let "S" denote the sum of the two numbers shown on the dice and let "K" = the number shown on the first die.

    -a) Determine whether the event S=8 is independent of K by computing the conditional probabilities P(S=8|K) for various vales of K and also comparing to P(S=8)

    -b)The same for S=10

    ----------------------------------------
    6. There are respectively 35, 25, and 15 students in the traditional, conceptual, and bridge geometry sections taught by three teachers, On Pat's birthday, the three sections congregate together for a celebration.

    -a) A student is randomly selected from the crowd of students. What is the expected number of students in the same section of this student?

    -b) If one of the three teachers is selected at random, what is the expected number of students taught by this teacher?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by JSharpie View Post

    1. A bakery has 100 bags of cookies in stock. Each bag contains 14 chocolate chip cookies. The probability that a cookie have fewer than 10 chips in it is 10%.
    Therefore:
    100 bags
    14 cookies in each bag
    =
    1400 total cookies
    140 cookies with less than 10 chips

    -a)Find the probability in a bag that there are exactly 3 cookies that have fewer than 10 chips in them.

    -b)What is the expected number of cookies, that have fewer than 10 chips, in a bag?

    -c)What is the standard deviation of the number of cookies, that have fewer than 10 chips, in a bag?

    -d) How many of the 100 bags of cookies would you expect to have exactly 3 cookies that have fewer than 10 chips?

    -e) What is the probability that between 14 and 25 (inclusive) bags out of the 100 bags have exactly 3 cookies with fewer than 10 chips?
    These are a binomial distribution problems.

    In the binomial distribution we have the probability of exactly n favourable
    occurrences in m trials is:

    <br />
P(n,m) = \frac{m!}{(m-n)!\ n!} p^n (1-p)^{m-n}<br />

    Where p is the probability of a favourable outcome on a single trial.

    In (a) the favourable outcome (from the point of view of doing the
    calculation not of the cookie eater) is that a cookie has fewer than 10
    chocolate chips, and a bag constitutes 14 trials, so in this case p=0.1,
    m=14 and n=3.

    The mean number of favourable outcomes in m trails for a
    binomial distributed random variable is:

    m(p,m)=p\ m

    which should allow you to do part (b).

    The standard deviation of the number of favourable outcomes in m trails
    for a binomial distributed random variable is:

    \sigma(p,m)=\sqrt{m\ p\ (1-p)}

    which should allow you to do part (c).

    (d) and (e) are further exercises on the binomial distribution but now
    our favourable outcome is that a bag contains exactly 3 cookies with 10
    or fewer choc chips, which is what you will have calculated in part (a).

    RonL
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  3. #3
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    Thanks, I still do not understand d and e. And the rest of the problems.
    So if it is correct it would be
    a) 11.42%
    b) 1.4
    c) 1.12

    Or am I still doing something wrong?
    Sorry if I sound dumb and hasty, I guess I am but I need to know all of this by monday. I'll send you x-mas cookies if I get helped.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by JSharpie View Post
    Thanks, I still do not understand d and e. And the rest of the problems.
    So if it is correct it would be
    a) 11.42%
    b) 1.4
    c) 1.12

    Or am I still doing something wrong?
    These all look right to me.

    d) asks out of 100 bags how many would you expect to have exactly 3
    cookies with 10 or fewer chips, this is 100x0.1142=11.42. This is 100 times
    the probability worked out in part a).

    e) What is the probability that between 14 and 25 (inclusive) bags out of the
    100 bags have exactly 3 cookies with fewer than 10 chips?

    This is:

    P=sum(n=14 ..25) 100!/[(100-n)! n!] (0.1142)^n (1-0,1142)^(100-n)

    which if you have a suitable calculator you can evaluate as it stands.

    However you are probably expected to use the normal approximation
    to the binaomial distribution for this.

    For this the mean is 11.42, and the standard deviation is

    sqrt[100x0.1142x(1-0.1142)] ~= 3.18.

    In this approximation we replace the range 14-25 by the continuity corrected range 13.5 - 25.5, which are

    (13.5 - 11.42)/3.18 ~=0.654,

    and

    (13.5 - 11.42)/3.18 ~=4.44

    standard deviations from the mean.

    Therefore the required probability is:

    P(4.44) - P(0.654) ~= 1 - 0.744 ~= 0.256.

    RonL
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  5. #5
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    Thanks, now any help on the others? Other than number 3, since I figured that one out.
    2, 4, 5, and/or 6?
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  6. #6
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    Re: Prob Homework help any of these 6 questions

    That kind of homework really sucks, I hate all of those 6 question. Grrrr... I don't want to study anymore.


    "I want to have EMT License to be part of rescue unit."
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