# Normal Distribution Porbability

I have ##\bar{X}## ~ ##N(\mu , 9/25)##

I have ##E[X] = \mu##
##Var[X] = 9/25##
##SD[X] = 3/5 = 0.6##

An interval for ##\bar{X}## has been recorded: ##\bar{X} \pm 1.05##.

I asked to find ##P(\bar{X} > \mu + 1.05)##

I can "normalize" the distribution through:

##Z = \frac{\bar{X} - \mu}{0.6}## ~ ##N(0,1)##

I'm confused by this next step:

##P(\bar{X} > \mu + 1.05) = P(Z > \frac{1.05}{0.6} = 1.75)##

I'm not sure how you go from the first probability to the other. Could any help please?

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Ray Vickson
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I have ##\bar{X}## ~ ##N(\mu , 9/25)##

I have ##E[X] = \mu##
##Var[X] = 9/25##
##SD[X] = 3/5 = 0.6##

An interval for ##\bar{X}## has been recorded: ##\bar{X} \pm 1.05##.

I asked to find ##P(\bar{X} > \mu + 1.05)##

I can "normalize" the distribution through:

##Z = \frac{\bar{X} - \mu}{0.6}## ~ ##N(0,1)##

I'm confused by this next step:

##P(\bar{X} > \mu + 1.05) = P(Z > \frac{1.05}{0.6} = 1.75)##

I'm not sure how you go from the first probability to the other. Could any help please?
The events ##\{\bar{X} > \mu + 1.05\}## and ##\{Z > 1.05/0.6\}## are the same:
$$\{ \bar{X} > \mu + 1.05\} = \{\bar{X} - \mu > 1.05 \} = \{ (\bar{X} - \mu)/0.6 > 1.05/0.6 \}$$

Last edited:
1 person
The events ##\{\bar{X} > \mu + 1.05\}## and ##\{Z > 1.05/0.6\}## are the same:
$$\{ \bar{X} > \mu + 1.05\} = \{\bar{X} - \mu > 0.15\} = \{ (\bar{X} - \mu)/0.6 > 1.05/0.6 \}$$
Oh I see, I was being a bit silly there, thanks!