Question

Given normal random variable Z , with mean μ=17.1 and standard deviation σ=6.2 , we obtain a sample of Z of size 60 . Consider the new random variable Xˉ which is the average of our sample.
Find the interval [μXˉ−a,μXˉ+a] symmetric about μXˉ satisfying P(Xˉ∈[μXˉ−a,μXˉ+a])=0.95

Answer 1

what's the distribution of the sample mean for a normal r.v.?

Answer 2

i dont for sure

Answer 3

\[X^{-} \sim N \left( \mu,\frac{ \sigma^2 }{n } \right)\]

Answer 4

meam is 17.1?

Answer 5

is \(X^-\) supposed to be \(\bar{X}\)?

Answer 6

yes

Answer 7

\bar{X} in latex

Answer 8

\overline{X} may work better, let's see
\(\overline{X}\)

Answer 9

ok

Answer 10

\[\frac{ \overline{X}-\mu }{ \Large\frac{ \sigma }{ \sqrt{n} } }\]is a standard normal

Answer 11

ok

Answer 12

mu=17.1, sigma = 6.2 and n=60
does that men x= .95?

Answer 13

\[\mu \overline{X}+a=18.6688...\]

Answer 14

\[\mu \overline{X} - a = 15.5311...\]

Answer 15

ok?

Answer 16

i think your \(\mu\overline{X}\) is really \(\mu_{\overline{X}}\)