Question

Can someone help with this question regarding normal distribution. If D~N (250,400) what is P(D<242 ∩ D>256)?

Answer 1

Hello!
You've posted If D~N (250,400) . This states that in a certain normal distribution, the mean is 250 and the standard deviation is 400. Note that 400 is 'way too large to be a proper std. dev. in this problem. Ensure that you've copied down the problem correctly.

Answer 2

@mathmale, I have checked and I have copied it down correctly

Answer 3

So be it.

Can someone help with this question regarding normal distribution. If D~N (250,400) what is P(D<242 ∩ D>256)?

Would this problem be any easier for you if you were to find P(242<D<256 first?

Answer 4

typically \[X\sim N(\mu,\sigma^2)\]

is the notation used

Answer 5

400 is the variance

Answer 6

400 is not the variance. I believe the square root of 400 (ie. 20) is the variance.

Answer 7

@SolomonZelman, would you be able to help?

Answer 8

\[\sigma^2=400\]

\(\sigma=\sqrt{400}=20\) is the standard deviation

Answer 9

https://en.wikipedia.org/wiki/Normal_distribution#Notation

Answer 10

Also, I'd be surprised if "P(D<242 ∩ D>256)" was the question. you might want to check it too

Answer 11

Note that N(250,20) is appropriate; the standard deviation, 20, is considerably less than the mean, 250. So, if you encounter a normal distribution such as N(250,400), double check whether that 400 represents the variance. With all due respect to Zarkon, I have yet to see a valid probability and statistics distribution given as N(mean, variance); rather, I've seen only N(mean, std. dev.).

Can we now proceed to solve the given problem, assuming that the variance is 400 and the st. dev. is 20?

Answer 12

Almost all my books use N(mu, sigma squared)


and I have a lot of them

Answer 13

you tend to collect a lot of stat books when you have a Ph.D. in mathematical statistics

Answer 14

Re: "Also, I'd be surprised if "P(D<242 ∩ D>256)" was the question. you might want to check it too" oddly formed, but still valid, question.

The probability expressed above could be calculated as follows: 1-P(242<D<256.

Answer 15

Please see this. @Zarkon

Answer 16

The way the problem is written makes it trivial

how many numbers are less than 242 and greater than 256 at the same time

Answer 17

Yes, how would you do it @Zarkon
I agree with Zarkon that the notation is (mean, variance ie. s.d. squared)

Answer 18

I did not notice that; I agree with you (Zarkon) on that.

@chau88: Do your learning materials explicitly explain somewhere that N(250,400) represents mean=250, variance=400?

Answer 19

@mathmale I don't think there should be any doubts on that, that should be common mathematical statistical knowledge. If you google that, you'll find that all sources give this

Answer 20

Without doing any calculations, chau88, can you now state the value of "what is P(D<242 ∩ D>256)"?

Answer 21

No, that's why i posted the question in the first place

Answer 22

*

Answer 23

think about it. See Zarkon's comments: "Also, I'd be surprised if "P(D<242 ∩ D>256)" was the question. you might want to check it too" and "how many numbers are less than 242 and greater than 256 at the same time?"

Answer 24

I know. I tried doing P(D<242) + P(D>256), but I didn't get the right answer. @mathmale

Answer 25

Let's go to a real life example.

suppose we're researching the weights of males of age 21 and over.

Would it make sense to ask, "what's the probability that such males would turn out to weigh less than 100 lb and more than 250 lb?"

Answer 26

No, but I can't really help it if that is really the question asked, can I?

Answer 27

They gave the answer to be 0.727.

Answer 28

The point is, chau88, we are doing some sample problems in the hope that doing them (or failing to do them) will help you answer the question at hand. I did not accuse you of error in sharing the problem that you were given.

Answer 29

Ok sorry @mathmale

Answer 30

if that is the answer given then the original question was

\[P(D<242\cup D>256)\]

Answer 31

Notice that Zardon has used "U" (denoting UNION, not interstion) in his conjecture, above?

If D were to be a member of a set of integers, then very clearly no D could be smaller than 252 AND greater than 260 (or whatever) at the same time. Can you agree with that?

Answer 32

Ok, I understand. If so, how to do the union of the two?

Answer 33

But if the actual operator were

Answer 34

then I don't see a solution!

Answer 35

sure you do...it would be zero

Answer 36

impossible events have probability zero

Answer 37

you should proceed like you originally thought

\[P(D<242\cup D>256)=P(D<242)+P(D>256)\]

Answer 38

since the two events are mutually exclusive

Answer 39

I proposed a solution earlier.

I also provided an example: the distribution of weights of males over 21.
If you want to find the probability that a given male weighs less than 200 OR (not AND) or more than 300, you can find the P that he will weigh between 200 and 300 and then SUBTRACT that probability from 1.

Answer 40

Event 1: man weighs less than 200 lb
Event 2: man weighs more than 250 lb

Clearly, these two events are mutually exclusive and the P of the intersection is zero, as Z. has already explained.

Answer 41

but the P of the UNION of these two events is likely between 0 and 1.

Answer 42

Yes, thanks both. However, when I did the addition and the union, I get 0.345-0.382=-0.037, which is definitely wrong

Answer 43

add!

Answer 44

I have to agree with Zarkon: Even if you have the "answer," the two events listed in your homework problem are mutually exclusive and the P of those two events happening simultaneously is 0. Bum question.
Or perhaps it was meant to be a trick question.

Answer 45

yes...major typo on that question

Answer 46

Believe me, such things do happen.
I'd suggest you move on to your next HW question.

Answer 47

Haha, good point. ok thanks @mathmale @Zarkon