# Math Help - normal distribution

1. ## normal distribution

hi, in my lecture notes, it says:

given, standard deviation of 0.1
$P(Z< \frac{xbar -0.3}{0.1/\sqrt{n}}) = 0.95$
$\frac{xbar -0.3}{0.1/\sqrt{n}} = 1.645$

this is a dumb question but how did they get to the second step?

2. Originally Posted by Dgphru
hi, in my lecture notes, it says:

given, standard deviation of 0.1
$P(Z< \frac{xbar -0.3}{0.1/\sqrt{n}}) = 0.95$
$\frac{xbar -0.3}{0.1/\sqrt{n}} = 1.645$

this is a dumb question but how did they get to the second step?
By using tables or technology you should be able to establish that for a standard normal variate, Pr(Z < 1.645) = 0.95 (correct to 4 decimal places). I hope you see the connection.