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[SOLVED] -aux.06.normal distribution to standard distribution

karush

Well-known member
Jan 31, 2012
3,084
so for (a) (i) I used $\frac{x-\mu}{\sigma}=z
$\dfrac{0.70-0.76}{0.06}=-1 = a $ and $\frac{0.79-0.76}{0.06}=.5 = b$
ii $P(.70<X)$ z-table for $-1$ is $0.3413$ so $0.3413 + .500 = 0.8413$
$P(.70<X<.79)$ z-table for $.5$ is $0.1915$ so $0.3413+0.1915=0.5328$
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Looks good so far! (Sun)
 

karush

Well-known member
Jan 31, 2012
3,084
(b) (i)
View attachment 1139

(ii) z-table for $3\% \approx -1.88$

so $\frac{c-0.76}{0.06}=-1.88$ thus $c\approx 0.65 s$

my shaky attempt at this anyway(Wasntme)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Again, looks good! (Clapping)
 

karush

Well-known member
Jan 31, 2012
3,084