Normal distribution tables confusion

This might be a useless comment, but I'd look up Phi(2.46), Phi(2.47) and use linear interpolation. That is,

\Phi (2.468) \approx 0.2 \Phi (2.46) + 0.8 \Phi (2.47).

Or, equivalently (I hope!),

\Phi (2.468) = \Phi (2.46) + 0.8[ \Phi(2.47) - \Phi(2.46)].

Reply 2

generalebriety

This might be a useless comment, but I'd look up Phi(2.46), Phi(2.47) and use linear interpolation. That is,

\Phi (2.468) \approx 0.2 \Phi (2.46) + 0.8 \Phi (2.47).

Or, equivalently (I hope!),

\Phi (2.468) = \Phi (2.46) + 0.8[ \Phi(2.47) - \Phi(2.46)].

Instead of just reading the table?? Doesn't your table booklet have 4 figure references?

Reply 3

coffeym

Instead of just reading the table?? Doesn't your table booklet have 4 figure references?

Nope.

Reply 4

generalebriety

Nope.

Oh very interesting...I thought all exam boards had to have a certain level of consistency between each of their formulae booklets.

Reply 5

batboy

OP

I am writing the Cambridge International Exams A level mathematics. The book published by Cambridge does not have anything about interpolation and so on. They just say to use the table at the back of the book.

I am so confused.

Reply 6

batboy

I am writing the Cambridge International Exams A level mathematics. The book published by Cambridge does not have anything about interpolation and so on. They just say to use the table at the back of the book.

I am so confused.

Could you find me a copy of this table? I never used anything like it.

Reply 7

DFranklin

18

batboy

Take for instance I have to find this out: P(Z <= 2.468) where Z ~ N(0,1)

So I used the table at the back of my book...

I look for the 2.4 in the ''y axis of the table". Good - I found that.

Then, I try to find the 6 in the top row of the table. These two lines intersect to give me 0.9931

Now, I try to find the 8 in the ADD portion. So I get 2 on the 2.4 line.

So I did what my textbook told me to do: 0.9931 + 2 X 10^-8Are you sure that's what the textbook says to do? I think it's very unlikely that you are supposed to add 2e-8; as you've found, that is far too small to make a difference to your answer. Conversely, if you added 2e-4 you would end up with the expected 0.9933.

As GE says, I'd use linear interpolation myself here.

Reply 8

Totally Tom

18

Some graphical calculators have normal distributions programmed in, if you have a Ti-84 or something similar it should have it so you can find the probability for any amount of decimals.

Reply 9

Cille

3

With regard to A Level / Further A Level statistics, OCR and MEI(OCR) have normal distribution tables accurate to 3 decimal places. The EDEXCEL tables are only accurate to two decimal places. EDEXCEL have not required students to use linear interpolation since before Syllabus 2000.

So, for example, if Z = 2/3 , EDEXCEL would require you to use z = 0.67 but OCR/MEI would expect a student to use z = 0.667.

Reply 10

nollaig

With regard to A Level / Further A Level statistics, OCR and MEI(OCR) have normal distribution tables accurate to 3 decimal places. The EDEXCEL tables are only accurate to two decimal places. EDEXCEL have not required students to use linear interpolation since before Syllabus 2000.

So, for example, if Z = 2/3 , EDEXCEL would require you to use z = 0.67 but OCR/MEI would expect a student to use z = 0.667.

Bizarre, then, that I was taught linear interpolation twice over the last two years, and asked to use it for every single stats homework and exam, and I did my exams with Edexcel...

Reply 11

generalebriety

Bizarre, then, that I was taught linear interpolation twice over the last two years, and asked to use it for every single stats homework and exam, and I did my exams with Edexcel...

I have never once been taught linear interpolation...I had better read up on it!

Reply 12

coffeym

I have never once been taught linear interpolation...I had better read up on it!

Heh. Nowt to read up on really. If you're asked to work out Phi(1.425), linear interpolation will tell you it's right between Phi(1.42) and Phi(1.43). If it's Phi(1.427) you need, just skew it slightly; take Phi(1.42) and add on 0.7 of the difference between Phi(1.42) and Phi(1.43), so you end up with Phi(1.427) = 0.7 Phi(1.43) + 0.3 Phi(1.42).

"Approximately equal to" signs where necessary, obviously.

Reply 13