Question

in the normal distribution N(4,4) find the probability \[P(X \ge 4.68)\]

Answer 1

Compute a Z-score and look up the tail probability in a table. Our Z-score is just$$Z=\frac{X-\mu}\sigma=\frac{4.86-4}4=\frac{0.86}4=0.24$$Typically, your table will tell you the *left* tail probability, i.e. \(P(Z\le0.24)\). Since we want the *right* tail probability, \(P(Z\ge0.24)\), just recall that the total probability must be \(1\):$$P(Z\le0.24)+P(Z\ge0.24)=1\\P(Z\ge0.24)=1-P(Z\le0.24)$$

Answer 2

Here's an example table for us to use:
http://intersci.ss.uci.edu/wiki/images/3/3a/Normal01.jpg

We see \(P(Z\le0.24)=0.5948\), so \(P(Z\ge0.24)=1-0.5948=0.4052\)

Answer 3

I see, so we look up in the table for the value of 0.24 and we find 0.5948 as our table value but once put into the equation it comes to
\[P(Z \ge 0.24) = 1 - 0.5948 = 0.4052\]
the answer in this case being 0.4052, is there a step after that?
because the answer on the answer sheet has 0.3669

Answer 4

also in your first step, I think you switched the numbers around :P
it's 4.68 not 4.86

Answer 5

but when I put in 4.68 it still doesn't come up as the requested answer @oldrin.bataku

Answer 6

It turns out I forgot that \(N(4,4)\) means \(\mu=4,\sigma^2=4\implies \sigma=2\). Sorry about that! Let's try again:$$Z=\frac{4.68-4}2=\frac{0.68}2=0.34$$We find \(P(Z\le0.34)=0.6331\) in our table and thus \(P(Z\ge0.34)=1-0.6331=0.3669\)

Answer 7

oh don't be sorry! I'm glad for your help!

now I understand, so \[N(4,40) \rightarrow \mu =4,\sigma ^{2} , \rightarrow \sigma = 2\] ?

Answer 8

thank you again @oldrin.bataku !!

Answer 9

\(N(4,4)\) means normal distribution with mean \(\mu=4\) and variance \(\sigma^2=4\). We're interested in the standard deviation \(\sigma\), though, which is just \(\sigma=\sqrt4=2\). The idea is that every normal distribution has the same shape, only translated around and/or uniformly stretched or compressed horizontally. Our standard normal distribution is the distribution centered at the origin with a standard deviation of \(1\), \(N(0,1)\). Because of this similarity, the idea is that you only need a table for the probabilities for the standard normal distribution and we can just scale/shift back and forth. Below is the distribution \(N(4,4)\).