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    I would like to know if my method is correct for 6(b). I'm fine with 6(a).

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    I did a quick sketch and realised that I have an area of 0.6 'at the right' of the normal distribution. Therefore, the value of z I need would be obtained by using the value of p=0.4.

    Using the curve symmetry, the z value at 0.4 would be the same as the z value at 0.6, but negative. Using Table 4 of the AQA formula book, it is -0.2533.

    I then plug this and the values given into the z-score standardisation formula (the one for the sample mean distribution), and rearranged to find an x of 19.4934.

    ...

    Is there a better approach that is more intuitive? It seems you could only do it this way by carefully looking at sketches you make. I'm not sure if my working is even correct despite getting the correct answer.
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    (Original post by W. A. Mozart)
    I would like to know if my method is correct for 6(b). I'm fine with 6(a).

    Name:  IMG_4361.jpg
Views: 13
Size:  505.7 KB

    I did a quick sketch and realised that I have an area of 0.6 'at the right' of the normal distribution. Therefore, the value of z I need would be obtained by using the value of p=0.4.

    Using the curve symmetry, the z value at 0.4 would be the same as the z value at 0.6, but negative. Using Table 4 of the AQA formula book, it is -0.2533.

    I then plug this and the values given into the z-score standardisation formula (the one for the sample mean distribution), and rearranged to find an x of 19.4934.

    ...

    Is there a better approach that is more intuitive? It seems you could only do it this way by carefully looking at sketches you make. I'm not sure if my working is even correct despite getting the correct answer.
    That appears to be the appropriate method - not sure about AQA in particular but it helps to be familiar with the sketch of the normal distribution and how the z values are related.
 
 
 
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