# Thread: probability and random samples

1. ## probability and random samples

Hi I have some uni problems in my exam revision i can't sort out.
m=215mg
sd=15mg
normal distribution
215mg of nicotine in a standard cigarette

a) find the probability a single cigarette has a nicotine content more than 220mg

b) if a random sample of 25 cigarettes is selected, what is the probability that the mean is more than 220mg

I tried using Z=y-u/sd but the z score from the table was 0.629 which can be correct right?
PLEASE help i dont know what model i should be using!!

Also, i have no idea about part b! HELP!

2. Originally Posted by kate204
Hi I have some uni problems in my exam revision i can't sort out.
m=215mg
sd=15mg
normal distribution
215mg of nicotine in a standard cigarette

a) find the probability a single cigarette has a nicotine content more than 220mg

b) if a random sample of 25 cigarettes is selected, what is the probability that the mean is more than 220mg

I tried using Z=y-u/sd but the z score from the table was 0.629 which can be correct right?

Mr F says: Obviously NOT since ${\color{red}220 > \mu = 215}$.

PLEASE help i dont know what model i should be using!!

Also, i have no idea about part b! HELP!
The model you're using is plainly the following:

X ~ Normal( $\mu = 215, ~ \sigma = 15$) where X is the random variable nicotine content of a single cigarette.

(a) $\Pr(X > 220) = \Pr \left( Z > \frac{220 - 215}{15} \right)$.

(b) You need to know the distribution of the sample mean:

$\overline{X}$ ~ Normal $\left( \mu = 215, ~ \sigma = \frac{15}{\sqrt{25}}\right)$

assuming an 'infinite' population of cigarettes. You must have a theorem in your textbook or class notes that shows where this has come from.

Now find $\Pr (\overline{X} > 220) = \Pr(Z > .......)$.

3. Originally Posted by kate204
Hi I have some uni problems in my exam revision i can't sort out.
m=215mg
sd=15mg
normal distribution
215mg of nicotine in a standard cigarette

a) find the probability a single cigarette has a nicotine content more than 220mg

b) if a random sample of 25 cigarettes is selected, what is the probability that the mean is more than 220mg

I tried using Z=y-u/sd but the z score from the table was 0.629 which can be correct right?
PLEASE help i dont know what model i should be using!!

Also, i have no idea about part b! HELP!

(a) $z=\frac{220-215}{15}=0.3333..$

Now look this up to find $P(Z<0.3333)\approx 0.631$, so to within the limits of accuracy of your table your answer looks OK.

(b) for the second part you need to use the result that the SD of a sample mean is the SD of an individual divided by the square root of the sample size.

RonL

4. Originally Posted by CaptainBlack
(a) $z=\frac{220-215}{15}=0.3333..$

Now look this up to find $P(Z<0.3333)\approx 0.631$, so to within the limits of accuracy of your table your answer looks OK.
[snip]
Sorry to be disagreeable but the answer is not OK. Not as a final answer anyway, which is how it was presented ....

The question asks "find the probability a single cigarette has a nicotine content MORE than 220mg" so the answer is 1 - 0.631 ......

5. sorry im still a little confused,
why is the final answer 1-0.631 instead of 0.631?

6. Originally Posted by kate204
sorry im still a little confused,
why is the final answer 1-0.631 instead of 0.631?
$\Pr (Z < 0.3333) \approx 0.631$. But the question is not asking this.

It should be clear (and my posts make this abundantly clear) that the question wants $\Pr(Z > 0.3333)$ .....

As I said at the start, 220 > 215 (the mean) ...... it must be obvious then that Pr(X > 220) cannot possibly be larger than 0.5 .....