# S1 Normal Distribution HelpWatch

#1
5. 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.
Assuming that the heights the high jumper can reach follow a Normal distribution,

(a) draw a sketch to illustrate the above information, (3)

(b) find, to 3 decimal places, the mean and the standard deviation of the heights the high jumper can reach, (6)

(c) calculate the probability that he can jump at least 1.74 m. (3)

Can someone explain in detail how I would attempt this question as I am getting no where

Thanks
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10 years ago
#2
you use the (x - mean)/st dev = z equation, and re-arrange for b, plug in what you know re arrange for what you don't. You will have 2 simultaneous equations to solve for the mean and st dev.

if you can't do part a or c, go through the book again as it means you don't understand how to do them at all.
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#3
I've got for part A the graph

Part B i got up to P(Z>1.78 - Mean / SD) = 0.2

And P(Z > 1.65 - Mean / SD) = 0,7

what i was confused on was that the mark scheme had the same as above but equal to 0.8416 and -0.5244 while these values are not on my s1 normal distribution table of values?

Part C i know what to do once i've found the mean and sd

sorry should of been more specific
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10 years ago
#4
Looking up 0.8 in the tables, you'll see that it occurs between the points 0.84 and 0.85. You want 5/28 of the way along to get 0.8. Since 5/28 = 0.18 approx, then linear interpolation gives you a z-value of 0.8418. Etc
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#5
why 0.8?
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#6
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10 years ago
#7
One in five is 20%. Hence for 80% of the time the jumper fails to reach that height: the area to the left of that height is 80%, or 0.8. Hence you need the z-value corresponding to 0.8. {You must always look up a "long tail" in normal tables. In this case, 0.8 is the long tail.}
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