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Maths - Statistics 1 Question

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Hey, does anyone know how to do this problem?

3. The distance in metres that a ball can be thrown by pupils at a particular school follows a normal
distribution with mean 35.0m and standard deviation 11.6 m.

(i) Find the probability that a randomly chosen pupil can throw a ball between 30 and 40 m. [3]

(ii) The school gives a certificate to the 10% of pupils who throw further than a certain distance.
Find the least distance that must be thrown to qualify for a certificate. [3]


...........I got the first part ,but dunno how to do the second..............
 
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P(X > d) = 0.1
P{Z > (d-35)/11.6} = 0.1
find the value of 0.9 from the table that is the inverse value then equate it to (d-35)/11.6
we will take 0.9 becoz it is greater than d and the table values are eligible for less than , use the critical values , these are given beneath the table in a smaller table, at the bottom of the page, the value for 0.9 is 1.282

1.282 = (d -35)/11.6
1.282 * 11.6 = d - 35
(1.282 * 11.6) + 35 = d
thus, d = 49.9m
==> the least distance is 49.9m
 
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one more question......how do you know when to use the critical values?
Is it whenever the value of probabilty is " 0.75, 0.90 ,0.95 ,0.975, 0.99, 0.995, 0.9975 ,0.999 ,0.9995" , you use the smaller table? ..........................I'm confused...... :?
 
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@the giver, u r spot on!

@ death valley:
its a bit tricky and confusing, its like this:
75% of the score is greater than 63
i'm taking standard deviation as 's'
P(X > 63)= 0.75
1 - P[Z < (63 - 51)/s]
by using the critical values, the value for 0.75 is 0.674
0.674 = (63-51)/s
0.674s = 12
s = 12/0.674
===> s =17.8
 
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