Normal distribution questions

[Yeti]

Not done any Stats for years and gotta have some work in by tomorrow.

In this question the mean is 218 and SD 10. I have to find the probability that an item selected at random is > 210.

Um, I have no idea.

[BruceZ]

[ QUOTE ]
Not done any Stats for years and gotta have some work in by tomorrow.

In this question the mean is 218 and SD 10. I have to find the probability that an item selected at random is > 210.

Um, I have no idea.

[/ QUOTE ]

78.8145% for being > than mean - 0.8 SD. In Excel, use 1 - NORMSDIST(8/10).

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Not done any Stats for years and gotta have some work in by tomorrow.

In this question the mean is 218 and SD 10. I have to find the probability that an item selected at random is > 210.

Um, I have no idea.

[/ QUOTE ]

78.8145% for being > than mean - 0.8 SD. In Excel, use 1 - NORMSDIST(8/10).

[/ QUOTE ]

Sorry, meant 1 - NORMSDIST(-8/10) since you are below the mean. That's what I computed, so the answer is still 78.8145%.

[dkernler]

[ QUOTE ]
In this question the mean is 218 and SD 10. I have to find the probability that an item selected at random is > 210.

[/ QUOTE ]
To give you a little context, you could first find a z-score. In general, z = (x - mean)/stdev. In this case, that's (210-218)/10 = -0.8. If you need to use a normal table, depending on your particular table, you then look up this z-score on the table. (In mine, the table gives the area up to the z-score, so I'd answer 1-0.2119 = 0.7881.)

Alternatively, you can use a TI calculator (or comparable) with the line normalcdf(210,1E99,218,10). That's under the [2nd] [DISTR] menu. It takes arguments (start,end,mean,stdev).

Of course, you can use Excel as well!

[Yeti]

Yeah, cheers guys! It's coming back to me, slightly.

This is a bit naughty of me but bear with me :

The next question involves a sample of 9 taken from the above question. I'm asked to find the probability that the total is between 1932 and 2022.

Do I divide this by 9 and work out the P( 214.67 < X < 224.67) or am I way off?

[elitegimp]

[ QUOTE ]
Yeah, cheers guys! It's coming back to me, slightly.

This is a bit naughty of me but bear with me :

The next question involves a sample of 9 taken from the above question. I'm asked to find the probability that the total is between 1932 and 2022.

Do I divide this by 9 and work out the P( 214.67 < X < 224.67) or am I way off?

[/ QUOTE ]

very close -- but when you multiply (or divide) a normal distribution, the variance / standard deviation change. So if you have a normal variable X with mean A and Standard Deviation B (variance B^2), then 9*X has a mean of 9*A and a Standard Deviation of [crap... I can't remember my signs! I think it's 3*B => variance of 9*B^2]. Easy enough to look up, but I don't have any stats books handy. Sorry.

[Yeti]

Ahah, thanks.

So,

We need to work out the P( 214.67 < X < 224.67) with SD of 10/3.

= P (-10/3 / 10/3) < Z < (20/3 / 10/3)
= P ( -1 < Z < 2)
= P (Z < 2) - P (Z < -1) = P (Z < 2) - (1 - P(Z < 1)
= 0.9772 - (1 - 0.8413)
= 0.8185

That right?

Unsure on the last 3 lines or so.

Thanks!

[BruceZ]

[ QUOTE ]
Ahah, thanks.

So,

We need to work out the P( 214.67 < X < 224.67) with SD of 10/3.

= P (-10/3 / 10/3) < Z < (20/3 / 10/3)
= P ( -1 < Z < 2)
= P (Z < 2) - P (Z < -1) = P (Z < 2) - (1 - P(Z < 1)
= 0.9772 - (1 - 0.8413)
= 0.8185

That right?

Unsure on the last 3 lines or so.

Thanks!

[/ QUOTE ]

That's correct. You can also work with the sum directly. The sum has a mean of 218*9 and a SD of 10*sqrt(9) = 10*3. To normalize (convert to a standard normal distribution with mean = 0 and SD = 1) subtract the mean and divide by the standard deviation:

(2022 - 9*218)/(10*3) = 2

(1932 - 9*218)/(10*3) = -1

NORMSDIST(2) - NORMSDIST(-1) = 81.86%.

Or, you can use the NORMDIST (without the S) function which takes the new mean and standard deviation into account for you:

NORMDIST(2022,9*218,3*10,TRUE) - NORMDIST(1932,9*218,3*10,TRUE)

= 81.86%.

Obviously this is a class problem, or it wouldn't have worked out to exactly 2 and -1 standard deviations.

[Yeti]

Ooh, that makes me happy. Thanks Bruce.

Last one :

What do I do when I have two groups of samples? They both have 16 members in each.

Two questions,

1) Prob that totals differ by > 210
2) Prob that the avg differs by > 2

Mean is still 218, SD 10.

I'll read the replies when I get up (in 6 hours, pah.) Thanks, I really appreciate this. Being on the last minute sucks.