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# S1 normal distribution question Watch

1. From experience a high jumper knows that he can clear a height of at least 1.78 m once in 5 attempts. He also knows that he can clear a height of at least 1.65 m on 7 out of 10 attempts.

Find, to 3 decimal places, the mean and the standard deviation of the heights the high jumper can reach

mark scheme: (1.78-u)/o^2 = 0.8416
(1.65-u)/o^2 = -0.52
44

can anyone tell me how the bolded numbers were obtained? i follow everything else

2. We did this the other day in class.

The mean is somewhere inbetween. So you know greater than 1.78 is 20% or 0.2. From this you can find the z value.

You also know less than 1.65 is 0.3. You can also find the z value for this.

From this you form a simultaneous equation and solve to find the standard deviation and the mean.
3. (Original post by Thrug)
We did this the other day in class.

The mean is somewhere inbetween. So you know greater than 1.78 is 20% or 0.2. From this you can find the z value.

You also know less than 1.65 is 0.3. You can also find the z value for this.

From this you form a simultaneous equation and solve to find the standard deviation and the mean.
how do we know that >1.78 is 20% and <1.65 is 30%?
4. (Original post by the maths guy)
how do we know that >1.78 is 20% and <1.65 is 30%?
It tells you in the question. You need to be able to pick up on "once in 5 attempts" is 1/5 and hence 0.2. Same rule applies for the 7/10. Except in this case greater than 1.65 is 0.7, so less than would be 0.3.

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Updated: December 8, 2010
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