I'm having trouble with this problem
any whelp will be appreciated
Where are you stuck? Have you been taught how to calculate probaiblities from a normal distribution? I'll do (b):
$\displaystyle Z = \frac{X - \mu}{\sigma}$.
z-score for 130: $\displaystyle z = \frac{130 - 172}{33} = -1.2727$.
z-score for 210: $\displaystyle z = \frac{2110 - 172}{33} = 1.1515$.
Therefore Pr(130 < X < 210) = Pr(-1.2727 < Z < 1.1515)
= Pr(Z < 1.1515) - Pr(Z < -1.2727)
= Pr(Z < 1.1515) - Pr(Z > 1.2727) by symmetry
= Pr(Z < 1.1515) - [1 - Pr(Z < 1.2727)] = Pr(Z < 1.1515) + Pr(Z > 1.2727) - 1.
Now use your tables to get the required probabilities.
well the thing is that the teacher just keep talking n talking n writing non-stop in the board all the time all i can do is copy the notes.
Also i got a question
For example:
Mean = 30 and standard deviation = 4
P(X<37) = P(Z> 37-30/4) = P(Z<1.75) = 0.5 + 0.4599 = 0.9537
So my question is where does that 0.5 come from that is also confusing me.
Also there is 2 different table i'm just not sure which one to use most of time
This is table 2
This is table 3
Thanks for your help
Table A-2 gives area to the left of the z-value, that is, Pr(Z < z) (but you have to add 0.5 to the value in the table - that's just how the table works).
Table A-3 gives area to the right of the z-value, that is, Pr(Z > z).
Using Table A-2: Pr(Z < 1.75) = 0.5 + 0.4599 = 0.9537. The 0.5 is added because that's how Table A-2 works.
Using Table A-3: Pr(Z < 1.75) = 1 - Pr(Z > 1.75) = 1 - (look it up in the table) = 0.9537.
It doesn't matter which table you use as long as you understand how to use it.