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    The weight, in grams, of beans in a tin is normallydistributed with mean 205 and standard deviation unknown .

    Given that 98% of tins contain between 200 g and 210 g find the value of the standard deviation.

    Okay so i've drawn out a diagram which I will attach incase it helps and I watched how the exam solutions guy did it and to solve for the standard deviation, he used the z value of 2.3263 for when the probability is 0.01 but he used the value of 210g which I don't understand because the area to the left of 210g isn't 0.01, it is 0.99?

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    the z value of 2.32... is used when p(z) is 0.99
 
 
 
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