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Math Help - Basic Probability Question Help

  1. #1
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    Basic Probability Question Help

    I need help with 2 questions:

    1) In a shipment of 18 trucks to a local truck dealer, there are 4 trucks that don't have air conditioning. Assume you select 4 trucks at random.

    What is the probability that two of the four don't have air conditioning?
    What is the probability that two or fewer of the four don't have air conditioning?

    I tried to use permutation for this problem, but I'm not sure if that is right or which numbers to use: n!/ (n-x)!


    2) The volume of a one-pound bag of coffee is normally distributed. Suppose you take a random sample of 15 one-pound bags of coffee. Find two values K sub L and K sub U such that the probability is 95% that the ratio of the sample standard deviation divided by the population standard deviation is between K sub L and K sub U.

    I don't understand what the question is asking or what K sub L and U are to be? Can both these questions be done on excel? If so, what formula?

    Thanks
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  2. #2
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    I got the first question by using hypergeometric distribution through excel.

    Still need help with the 2nd question. I don't understand the phrase "Find two values K sub L and K sub U such that the probability is 95% that the ratio of the sample standard deviation divided by the population standard deviation is between K sub L and K sub U."

    I was told to use confidence interval and chi square but not sure how to apply these
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  3. #3
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    JHello, armani!

    1) In a shipment of 18 trucks to a local truck dealer,
    there are 4 trucks that don't have air conditioning.
    Assume you select 4 trucks at random.

    \displaystyle \text{There are: }\;{18\choose4} = 3060\,\text{ possible outcomes.}




    (a) What is the probability that two of the four don't have A/C?

    We want 2 with AC and 2 without AC.

    \displaystyle \text{There are: }\;{14\choose2}{4\choose2} \:=\:546\text{ ways.}

    \text{Therefore: }\;P(2\sim\!\!\text{AC}) \;=\;\dfrac{546}{3060} \;=\;\dfrac{91}{510}




    (b) What is the probability that two or fewer of the four don't have A/C?

    \text{"Two or fewer without AC" means: }\:0 \sim\!\!\text{AC}\,\text{ or }\, 1\sim\!\!\text{AC}\,\text{ or }\,2\sim\!\!\text{AC}

    \text{The opposite is "3 or more without AC"}
    . . \text{which means: }\,3\sim\!\!\text{AC}\, \text{ or }\,4\sim\!\!\text{AC}


    . . P(3\sim\!\!\text{AC}) \;=\;\dfrac{{4\choose3}{14\choose1}}{3060} \;=\;\dfrac{56}{3060}

    . . P(4\sim\!\!\text{AC}) \;=\;\dfrac{{4\choose4}}{3060} \;=\;\dfrac{1}{3060}


    \displaystyle \text{Hence: }\;P(\text{3 or more}\sim\!\!\text{AC}) \;=\;\frac{56}{3060} + \frac{1}{3060} \;=\;\frac{57}{3060}


    \displaystyle \text{Therefore: }\;P(\text{2 or fewer}\sim\!\!\text{AC}) \;=\;1 - \frac{57}{3060} \;=\;\frac{3003}{3060} \;=\;\frac{1001}{1020}
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  4. #4
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    Thank you soroban

    Those are the exact same answers I got as well.

    I still do not know how quite to get the answer to question 2 by applying confidence interval. I'm not given a data set to find the sample mean so I cannot find the sample standard dev.

    any help on this?
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