Normal distribution Q10
I didn't know how to do any of these.
For part i) i tried:
X~ N(1.34, 0.021)
P(x>1.37)
z=(1.37-1.34)/sqrt(0.021)=0.20702
so p(z> 0.20702) = 1- 0.5425 (from the normal distribution tables for 0.20702)
but it was wrong!!
answers are meant to be:
i) 0.077
ii) 0.847
iii) 0.674
iv)1.313m
Original post by 0utdoorz
I didn't know how to do any of these.
For part i) i tried:
X~ N(1.34, 0.021)
P(x>1.37)
z=(1.37-1.34)/sqrt(0.021)=0.20702
so p(z> 0.20702) = 1- 0.5425 (from the normal distribution tables for 0.20702)
but it was wrong!!
answers are meant to be:
i) 0.077
ii) 0.847
iii) 0.674
iv)1.313m
I didn't know how to do any of these.
For part i) i tried:
X~ N(1.34, 0.021)
P(x>1.37)
z=(1.37-1.34)/sqrt(0.021)=0.20702
so p(z> 0.20702) = 1- 0.5425 (from the normal distribution tables for 0.20702)
but it was wrong!!
answers are meant to be:
i) 0.077
ii) 0.847
iii) 0.674
iv)1.313m
Notice that, your square rooting in wrong, you've being told the standard deviation, the variable should be std deviation squared so X~N(1.34,0.021^2)
(edited 11 years ago)
Original post by Robbie242
Notice that, your square rooting in wrong, you've being told the standard deviation, the variable should be std deviation squared so X~N(1.34,0.021^2)
thanks! I've worked out part i,ii and iii now
how would I do part iv?
Original post by 0utdoorz
thanks! I've worked out part i,ii and iii now
how would I do part iv?
how would I do part iv?
Notice that 0.10000... is in the percentage points table, once you standardise using the z formula, try to inverse the z results (i.e. negative)
Original post by Robbie242
Notice that 0.10000... is in the percentage points table, once you standardise using the z formula, try to inverse the z results (i.e. negative)
by standardising using the z formula, do you mean p(x< (L-1.34)/0.021) =0.1?
Original post by 0utdoorz
by standardising using the z formula, do you mean p(x< (L-1.34)/0.021) =0.1?
Yup then make if that helps (find z on the percentage tables)
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