Normal Distribution Porbability

[Calu]

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?

 

[Ray Vickson]

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:
$$\{ \bar{X} > \mu + 1.05\} = \{\bar{X} - \mu > 1.05 \} = \{ (\bar{X} - \mu)/0.6 > 1.05/0.6 \}$$

 

[Calu]

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!