#### Monkeyseat

##### Full Member

- Joined
- Jul 3, 2005

- Messages
- 298

Question:

Working:10)

A survey showed that the value of the change carried by an adult male shopper may be modelled by a normal distribution with mean £3.10 and standard deviation £0.90. Find the probability that:

a) an adult male shopper selected at random will be carrying between £3 and £4 in change.

b) the mean amount of change carried by a random sample of nine adult male shoppers will be between £3.00 and £3.05

Give two reasons why, although the normal distribution may provide and adequate model, it cannot in these circumstances provide an exact model.

If you could check those for me that would be brilliant. Thanks. I thought it might be to do with the book using a more detailed version of the tables (although I used the one from the book) to further decimal places, i.e. being able to find a more accurate value when looking up -0.111111 etc. in the tables. In part (b) I didn't round originally and just looked up -0.16 but this was also wrong according to the book.10)

a)

z value = (X - mean)/standard deviation

z = (3 - 3.1)/0.9

z = -0.111111111 etc.

Looking up -0.11 in the table gives 0.54380. I need the probability of less than the negative z value, so I did 1 - 0.54380 = 0.4562.

z value = (X - mean)/standard deviation

z = (4 - 3.1)/0.9

z = 1

Again, from the tables, looking up 1 gives a probability of 0.84134.

Therefore; P(3 < X < 4) = 0.84134 - 0.4562

This gives0.38514. That is my final answer but the book says it is0.386... Who is correct?

----------------

b)

Where x bar is the sample mean (at least I think that's what it means from how I interpreted the book - the teacher missed the bar off altogether most of the time when learning about the central limit theorem so I'm not sure), n is the sample size and sqr. the square root sign:

z value = (x bar - mean)/(standard deviation/sqrt. n)

z = (3 - 3.1)/(0.9/sqrt. 9)

z = -0.333333333 etc.

When I looked up -0.33 in the tables it gave me the value 0.62930. The probability of less than the negative z value is needed, therefore 1 - 0.62930 = 0.3707.

z value = (x bar - mean)/(standard deviation/sqrt. n)

z = (3.05 - 3.1)/(0.9/sqrt. 9)

z = -0.16666666

My table only goes up to 2 d.p., so like I was told in the past, I rounded to -0.17. I looked up -0.17 in the tables and got 0.56749. Again, to get P(X < 3.05) you have to do 1 - 0.56749 which gives 0.43251.

Therefore, the probability that mean will be between 3 and 3.05 is; P(3 < M < 3.05) = 0.43251 - 0.3707

The answer to this is0.06181.The book says the answer is0.0644. Who is correct?

And by the way, for the last bit of the question about explaining why the normal distribution cannot in these circumstances provide an exact model, I'm not sure on this. I'm don't understand the normal distribution 100% yet as much as I'm trying. I put for this section that the sample size was small which my teacher also said was the main reason, but this only really applies to part (b) where the sample size is given doesn't it? Also, the book says something about not being able to have "negative amounts of change" and it being impossible - "money is discrete variable normal is continuous" it says, not entirely sure how that impacts the normal distribution though...? In the context of the question, what does this mean/how do you get a negative probability (is that the negative z value as shown in my answer?)? The teacher said the book was "clutching at straws" regarding the "negative change" reason.

What does it actually mean, "normally distributed?" I don't understand it exactly.

Sorry to post another normal distribution question but I have done many of these questions and have only posted 3 threads, I was just wondering about this question due to different answers in the back - on previous questions the book says the answers may be different due to "interpolation", however it doesn't say this for this question so I don't know whether it applies (I don't know what it means either ).

This post is a bit long, so thank you for your time, it is much appreciated. Typo free hopefully hehe...

Anyway, cheers.