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    Can someone explain why they said that p(x>15)=p(y<or equal to 4) its the last line on the workout example. part a
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    (Original post by assassinbunny123)
    Can someone explain why they said that p(x>15)=p(y<or equal to 4) its the last line on the workout example. part a
    Because having more than 15 customers buy the tea is the same as having less than or equal to 4 customer NOT buying the tea.

    Reverse the logical statement.
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    (Original post by assassinbunny123)
    Can someone explain why they said that p(x>15)=p(y<or equal to 4) its the last line on the workout example. part a
    There are two ways to work out the answer. Either by considering the prob of buying tea =0.7 or by looking at fewer than 5 not buying tea. I think most people would go for the first approach but they are just saying there is another way.
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    (Original post by Muttley79)
    There are two ways to work out the answer. Either by considering the prob of buying tea =0.7 or by looking at fewer than 5 not buying tea. I think most people would go for the first approach but they are just saying there is another way.
    I don't know too much about old spec stats but wouldn't you have to use the Y random variable since the probabilities in the table are < 0.5?

    In the new spec you'd just use your calculator with p = 0.7.
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    (Original post by Notnek)
    I don't know too much about old spec stats but wouldn't you have to use the Y random variable since the probabilities in the table are < 0.5?

    In the new spec you'd just use your calculator with p = 0.7.
    I did not know whether OP is old or new spec.
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    (Original post by Muttley79)
    I did not know whether OP is old or new spec.
    I think it's an old spec textbook but the OP could be using it for new spec.
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    (Original post by Notnek)
    I think it's an old spec textbook but the OP could be using it for new spec.
    Lots of schools are using the old books.
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    (Original post by Muttley79)
    Lots of schools are using the old books.
    Yes they are which is a shame, especially for stats.
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    (Original post by Notnek)
    Yes they are which is a shame, especially for stats.
    I agree...
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    (Original post by Notnek)
    I don't know too much about old spec stats but wouldn't you have to use the Y random variable since the probabilities in the table are < 0.5?

    In the new spec you'd just use your calculator with p = 0.7.
    old spec but how would you do this question on a calculator since i find going to the tables a bit tedious
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    (Original post by assassinbunny123)
    old spec but how would you do this question on a calculator since i find going to the tables a bit tedious
    Probability P(X=k) for a binomial distribution X \sim B(n,p) is \displaystyle \binom{n}{k} p^k (1-p)^{n-k}. So for P(X \leq k) you simply add up all the 'equality probabilities' P(X=0) + P(X=1) + ... + P(X=k), which can be denoted by \displaystyle \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}.

    In your question, you are looking for P(X &gt; 15) = P(X \geq 16) from X \sim B(20,0.7) which can be found from the calculator by typing in \displaystyle \sum_{i=16}^{20} \binom{20}{i} 0.7^i \cdot 0.3^{20-i} \approx 0.2375

    If you calculator doesnt have a sum button, then you'd need to sum these probabilities individually, which as you might expect is tedious as well if you're summing up many of them - hence the table in the formula booklet.
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    (Original post by assassinbunny123)
    old spec but how would you do this question on a calculator since i find going to the tables a bit tedious
    You’re on old spec so probably don’t have a binomial distribution function in your calculator. You could sum the probabilities but it’s a waste of time.

    A lot of questions will assume you’re using tables so it’s best to stick to tables for old spec even if your calculator does have the function.
 
 
 
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