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# S1 Doubt

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1. So, I have a calculator that can find out values from a standard distribution graph without the need to look at the tables. My question is that will I lose one mark, if a question comes up like this :

Find P(Z < -0.2352941176)

For this you need to do do 1 - P(Z > 0.2352941176)

Instead I can input P(-0.2352941176) into my calc directly and get the correct answer without needing to write the step in bold.

Will I lose a mark if I miss out the step in bold?
2. (Original post by Blazyy)
Will I lose a mark if I miss out the step in bold?
No, you won't.
3. (Original post by Zacken)
No, you won't.
Thanks a bunch, I have a few more doubts, do you mind?
4. (Original post by Blazyy)
Thanks a bunch, I have a few more doubts, do you mind?
5. (Original post by Zacken)
How do you do part b of this question?
http://prntscr.com/ba68ia
6. (Original post by Blazyy)
How do you do part b of this question?
http://prntscr.com/ba68ia
P(X < 1) = 0.05 ===> P(Z < (1 - mean)/standard deviation) = 0.05

And you know that P(Z < z) = 0.05 means that z = -1.6449 or w/e

So equate (1- mean)/s.d to -1.6449 and solve.
7. (Original post by Zacken)
And you know that P(Z < z) = 0.05 means that z = -1.6449 or w/e
I don't get this part
8. Edexcel explicitly say on te front of their papers that you will not get full marks for a question unless the method is shown.
9. (Original post by Blazyy)
I don't get this part
What don't you get? Your percentage points table shows that P(Z > 1.6449) = 0.05.

You have P(Z < z) = 0.05. Convert this to P(Z > -z) = 0.05. Then -z = 1.6449.
10. (Original post by Zacken)
What don't you get? Your percentage points table shows that P(Z > 1.6449) = 0.05.

You have P(Z < z) = 0.05. Convert this to P(Z > -z) = 0.05. Then -z = 1.6449.
Okay I understand, thanks.
11. The standard normal distributions give values for P(Z<z) where z>0. The probability distribution is symmetric about z=0 so we only need half the values.
If you want P(Z<z) = 0.05, then z must be negative. Sketch the distribution to see this. Then
P(Z<z) = 0.05 (where z is negative) is the same as P(Z>z) = 0.05 (z is now the positive value) which is the same as 1 - P(Z<z) where z is the positive value. My advice is to keep sketching the distribution for each question, shade in the bits for which you are given information and eventually you will be able to do the questions without a graph.
12. (Original post by Blazyy)
Okay I understand, thanks.
No problem.

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