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# Normal distribution Q7 Watch

1. I'm stuck on part ii

for part i) my answer is 78.65%

for part ii) I tried to do this:

p(x<5.2) = 0.16
p(z< (5.2-mean)/SD)=0.16

using inverse normal distribution tables for 0.84 (since 1-0.16=0.84)

5.2-mean/SD = 0.9945
5.2-mean= 0.9945*SD
5.2= mean + 0.9945*SD

and

p(x>5.3)=0.2
p(z>(5.3-mean)/SD)=0.2

using inverse tables for 0.8:

5.3-mean/SD = 0.8416

5.3=mean+ 0.8416*SD

then I did simultaneous equations:

5.3= mean + 0.8416*SD
5.2=mean + 0.9945*SD

which makes
0.1 = -0.1529*SD

but the answer I got for the Standard deviation was wrong so I couldn't find the mean.

the answers are meant to be 5.254 and 0.054
2. You have P(x<5.2) = 0.16. You want the o.16 to be greater than 0.5000 (so you can read from the table). So you have P(x < 2mean - 5.2) = 0.84. You ca normalise this and then use simultaneous equations with the rest - which looks fine (although I don't know how you got 0.8416- I would just use the closest value of Z that gives a probability of 0.8).

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Updated: April 17, 2013
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