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Math Help - probability

  1. #1
    amber79
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    probability

    hey all can anyone help me with these problems?

    1)a die is tossed 150 times, the rolling of a 6 is counted as a success:

    a) find the mean and standard deviation of this binominal experiment and

    b) what is the probability of throwing between 30 and 35 sixes?

    2) the mean weight of a group of 200 people is 74kg. the standard deviation is 9kg. what is the probability that a person selected at random weighs:

    a)between 90 and 100kg?
    b)between 70 and 100kg?
    c)is over 101kg?
    d)weighs less than 65kgs?
    e)how many in the group would weigh less than 65kgs?

    any help will be appreciated.....thanks
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  2. #2
    Grand Panjandrum
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    You should consider asking only one question in a single post
    it makes it easier to tell what has been answered, as it is
    not always possible to answer multiple questions in one go

    Quote Originally Posted by amber79
    hey all can anyone help me with these problems?

    1)a die is tossed 150 times, the rolling of a 6 is counted as a success:

    a) find the mean and standard deviation of this binominal experiment and

    b) what is the probability of throwing between 30 and 35 sixes?
    Part a)
    Assume a fair dice, the the probability p of success is 1/6.

    In N trials the mean number of successes is:

    m=p.N=150/6=25.

    In a binomial experiment with probability of success p and N
    trials the standard deviation of the number of success is:

    \sigma=\sqrt{N.p.(1-p)}=\sqrt{150.\frac{1}{6} .\frac{5}{6}}\approx 4.56

    Part b)
    If we use the normal approximation we want to find the probability of a
    normally distributed random variable X with mean 25 and
    standard deviation \approx 4.56 taking values between 29.5 and 35.5
    (we extend the range from (30, 35) to (29.5,35.5) as a continuity
    correction) or:

    prob(29.5<X<35.5) =  P((35.5-25)/4.56)-P((29.5-25)/4.56),

    where P denotes the cumulative distribution function for the
    standard normal distribution.

    prob(29.5<X<35.5) =  P(2.30)-P(0.99)\approx 0.9893-0.8389=0.1504 \approx 15 \%.

    RonL
    Last edited by CaptainBlack; January 2nd 2006 at 09:27 AM.
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  3. #3
    Global Moderator

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    If I make a mistake in how to slove the problem do not blame me I am not familiar with Probabilty and Statistics but I think I might be able to help with some of them.

    Part 2) the mean weight of a group of 200 people is 74kg. the standard deviation is 9kg. what is the probability that a person selected at random weighs:

    a)between 90 and 100kg?
    b)between 70 and 100kg?
    c)is over 101kg?
    d)weighs less than 65kgs?
    e)how many in the group would weigh less than 65kgs?


    I do not know a or b I could approximate what the answer is but I do not know if that is what the problem is asking for.

    d)There is a rule in statistics the first deviation Gives 68% two Deviations Give 95% and three give 99.7%. Since 74 is the mean one standard deviation over is 74-9 and 74+9 which is 65 and 83. Thus, 68% of the people are between 65 and 83. Thus are are 32% left. Which means there are 32% of people below 65 or more the than 83. But the Standard Deviation curve is symetrically, thus 16% are below 65 and 16% are above 83. But the problem is asking for below 65 thus the answer is 16%.
    e) Since 16% of the people are below 65 and there are 200 we have 16% of 200 = 32 people.
    c)Same reasoning as in part c this is why I did it out of order. Between 65 and 83 you have 68% of all the people. Move another 9 units (because that is the standard deviation) to get 65-9 and 83+9 thus 56 and 92 which means between 56 and 92 there are 95% of the people. Move the standard deviation for the third time thus 56-9 and 92+9 which is 47 and 101 which means between 47 amd 101 there are 99.7% of all the people. Thus .3% remains which means .3% of all the people are below 47 or above 101. But the standard deviation curve is symetrically thus .15% are above 101 and .15% are below 101. The problem is asking for above 101 thus the answer is .15%.
    Last edited by ThePerfectHacker; January 2nd 2006 at 11:50 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by amber79
    2) the mean weight of a group of 200 people is 74kg. the standard deviation is 9kg. what is the probability that a person selected at random weighs:

    a)between 90 and 100kg?
    b)between 70 and 100kg?
    c)is over 101kg?
    d)weighs less than 65kgs?
    e)how many in the group would weigh less than 65kgs?
    Here you are required to assume that weight is normally distributed, and
    as 200 is a reasonably large number in this context we may use 74kg as the
    mean and 9kg as the standard deviation of the weight distribution.

    Then if P is the cumulative standard normal distribution:

    prob(weight<x)=P\left ( \frac{x-mean}{standard\ deviation}\right)

    prob(weight>x)=1-P\left ( \frac{x-mean}{standard\ deviation}\right)


    Also:

    prob(y<weight<x)=P\left ( \frac{x-mean}{standard\ deviation}\right) -\ P\left ( \frac{y-mean}{standard\ deviation}\right)

    P is what is tabulated in a normal probability distribution table.

    With the above you should be able to answer a-e by plugging in the relevant
    values given in the problem.

    RonL
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