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    It is given that Y ∼ N(33, 21). Find the value of a given that P(33 − a < Y < 33 + a) = 0.5.?

    Please explain thoroughly. I do not get it , any kind of help would be great
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    (Original post by Sammysammy99)
    It is given that Y ∼ N(33, 21). Find the value of a given that P(33 − a < Y < 33 + a) = 0.5.?

    Please explain thoroughly. I do not get it , any kind of help would be great
    First of all, draw a nice sketch of the normal distribution for Y, marking on the mean.

    You should notice that the region 33-a &lt; Y &lt; 33+a is actually symmetrical about the mean, which means that the probabilities P(33-a &lt; Y &lt; 33) and P(33&lt;Y&lt;33+a) are indeed the same.

    This means we only need to focus on one of them if we can.
    Indeed we can because know that

    \begin{aligned} P(33-a &lt; Y &lt; 33+a) & = P(33-a &lt; Y &lt; 33) + P(33 &lt; Y &lt; a+33) \\ & = 2P(33 &lt; Y &lt; 33+a) = 0.5

    This implies that we have P(33 &lt; Y &lt; 33+a) = 0.25.

    We can also rewrite the LHS by saying that P(33 &lt; Y &lt; 33+a) = P(Y&lt;33+a) - P(Y &lt; 33) hence we have P(Y&lt;33+a) - 0.5 = 0.25 hence P(Y &lt; 33+a) = 0.75

    From here it should be straight-forward to determine a such that P(Y &lt; 33+a) = 0.75 holds.
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    (Original post by RDKGames)
    First of all, draw a nice sketch of the normal distribution for Y, marking on the mean.

    You should notice that the region 33-a &lt; Y &lt; 33+a is actually symmetrical about the mean, which means that the probabilities P(33-a &lt; Y &lt; 33) and P(33&lt;Y&lt;33+a) are indeed the same. This means we only need to focus on one of them if we can.
    Indeed we can because know that P(33-a &lt; Y &lt; 33+a) = P(33-a &lt; Y &lt; 33) + P(33 &lt; Y &lt; a+33) = 2P(33 &lt; Y &lt; 33+a) = 0.5

    This implies that we have P(33 &lt; Y &lt; 33+a) = 0.25. We can also rewrite the LHS by saying that P(33 &lt; Y &lt; 33+a) = P(Y&lt;33+a) - P(Y &lt; 33) hence we have P(Y&lt;33+a) - 0.5 = 0.25 hence P(Y &lt; 33+a) = 0.75

    From here it should be straight-forward to determine a such that P(Y &lt; 33+a) = 0.75 holds.
    Thank you very much , I really appreciate all that , I get it , but the only question is that is p(y<33) equal to 0.5? Thank u again , really helpful
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    (Original post by Sammysammy99)
    Thank you very much , I really appreciate all that , I get it , but the only question is that is p(y<33) equal to 0.5? Thank u again , really helpful
    Because 33 is the mean, and in normal distribution the mean=mode=median.
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    (Original post by Sammysammy99)
    Thank you very much , I really appreciate all that , I get it , but the only question is that is p(y<33) equal to 0.5? Thank u again , really helpful
    Because 33 is the mean, and we know that normal distribution is symmetrical about the mean which means that the probability is 50% on either side of the mean.
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    (Original post by Radioactivedecay)
    Because 33 is the mean, and in normal distribution the mean=mode=median.
    I just meant where did u get the 0.5 from
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    (Original post by Sammysammy99)
    I just meant where did u get the 0.5 from
    From the 50% I'm talking about.
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    (Original post by Sammysammy99)
    I just meant where did u get the 0.5 from
    Think about the shape of the distribution. It is symmetrical about the mean with 0 skew, hence the P(X>33)=P(X<33)=0.5
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    (Original post by RDKGames)
    First of all, draw a nice sketch of the normal distribution for Y, marking on the mean.

    You should notice that the region 33-a &lt; Y &lt; 33+a is actually symmetrical about the mean, which means that the probabilities P(33-a &lt; Y &lt; 33) and P(33&lt;Y&lt;33+a) are indeed the same.

    This means we only need to focus on one of them if we can.
    Indeed we can because know that

    \begin{aligned} P(33-a &lt; Y &lt; 33+a) & = P(33-a &lt; Y &lt; 33) + P(33 &lt; Y &lt; a+33) \\ & = 2P(33 &lt; Y &lt; 33+a) = 0.5

    This implies that we have P(33 &lt; Y &lt; 33+a) = 0.25.

    We can also rewrite the LHS by saying that P(33 &lt; Y &lt; 33+a) = P(Y&lt;33+a) - P(Y &lt; 33) hence we have P(Y&lt;33+a) - 0.5 = 0.25 hence P(Y &lt; 33+a) = 0.75

    From here it should be straight-forward to determine a such that P(Y &lt; 33+a) = 0.75 holds.
    Alright thank you very muchhhhh I get it now!!!
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