Question 921404: Hello all, i have a question about normal distrubution.
An archer fires arrows at a target (bulls eye) one thousand times. As- sume his hits and misses are normally distributed, with zero marking the bulls eye and a standard deviation of 10cm.
(i) How many arrows are within 1cm of the bulls eye?
(ii) If the circular board that contains the bulls eye at its centre is 1 meter wide, How many arrows land on the grass?
(iii) If another archer can hit within 1cm of the target 99% of the time, what is the standard deviation of his shots?
I understand how this question works, but i just wanna ask that without using the cumulative standard normal probability, can i still do this question with my Casio fx-991 ES plus. i found online that i can figure out the (phi z) with this calculator, but i do not know how to find the inverse for question part 3.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Hi,
Casio fx-991 ES plus... Mode(scroll down to the DIST mode)...
You will find 'Inverse Normal' among the choices.
•Distribution functions
(normal distribution and inverse normal distribution, binomial distribution, Poisson distribution)
(i) How many arrows are within 1cm of the bulls eye?  z = ± 1/10 , P(z <.1) - P(z < -.1) = .5398 - .4602
(ii) If the circular board that contains the bulls eye at its centre is 1 meter wide(100cm), How many arrows land on the grass?
z = ± 50/10
P(z <5) - P(z < -5) = .99999997 - .0000002867 Basically NO Arrows will fall in Grass
(iii) If another archer can hit within 1cm of the target 99% of the time, what is the standard deviation of his shots?
Inv z(.99)= 1/sd, sd = 1/Inv z(.99)
z= Inv z(.99) = 2.3263
Note:
Empirical Rule
one standard deviation from the mean accounts for about 68% of the set
two standard deviations from the mean account for about 95%
and three standard deviations from the mean account for about 99.7%. ***
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