Why do I get the answer 0?

najzub

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Joined
May 4, 2019
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4
Hi!

So here is the problem I'm working on: X is normally distributed with Expectation=2,1 and Standard Deviation=0,6. Why do I end up with the answer P(1,5≤X≤2,7)=0? Am I doing something wrong?
 
Yes.

We can't tell what you are doing wrong without seeing your work. Please show how you got your result, step by step. In particular, be sure to show the values you used for z, and tell us how you found probabilities.
 
Yes.

We can't tell what you are doing wrong without seeing your work. Please show how you got your result, step by step. In particular, be sure to show the values you used for z, and tell us how you found probabilities.

Certainly sir, thank you for your time. Please have a look at the picture below ( Calculation of P(X<1,5) and P(X>2,7) was also a part of the problem). I calculated z using the standard formula: (x-exp)/SD. Then I simply looked up the values for z in a table for cumulative standard normal distribution, which I used as shown in the picture.


11994
 
So here is the problem I'm working on: X is normally distributed with Expectation=2,1 and Standard Deviation=0,6. Why do I end up with the answer P(1,5≤X≤2,7)=0? Am I doing something wrong?
Here are housekeeping matters. Most all helpers on this site are from North America therefore we often misread posts. For example, you wrote 2,1 which is confusing because we are expecting \(\displaystyle 2.1\). I guesss that by exception the question implies mean.
If you notice that \(\displaystyle \sigma=0.6\), the S.D. so \(\displaystyle 1.5=2.1-\sigma<2.1+\sigma=2.7\).
So what percent is within one S.D. of the mean?
 
Certainly sir, thank you for your time. Please have a look at the picture below ( Calculation of P(X<1,5) and P(X>2,7) was also a part of the problem). I calculated z using the standard formula: (x-exp)/SD. Then I simply looked up the values for z in a table for cumulative standard normal distribution, which I used as shown in the picture.


View attachment 11994
The work is good until the final calculation. (Incidentally, I understood completely what you asked, in spite of some differences in terminology and notation; and here, it was very helpful that you showed all the details, including what you mean by G(z).)

The probability that you want is G(1) - G(-1), but you accidentally replaced G(1) with 1 - G(1), which you had calculated for the previous part of the problem!

The correct calculation is 0.8413 - 0.1587. Right?
 
Here are housekeeping matters. Most all helpers on this site are from North America therefore we often misread posts. For example, you wrote 2,1 which is confusing because we are expecting \(\displaystyle 2.1\). I guesss that by exception the question implies mean.
If you notice that \(\displaystyle \sigma=0.6\), the S.D. so \(\displaystyle 1.5=2.1-\sigma<2.1+\sigma=2.7\).
So what percent is within one S.D. of the mean?

Aaah, 68%! Right?..

EDIT: As Dr. Peterson pointed out, it seems like I actually made a mistake in one of those preceeding problems. This mistake was a win-win, because understood the concept of the 68-95-99.7 rule now 8-)
 
The work is good until the final calculation. (Incidentally, I understood completely what you asked, in spite of some differences in terminology and notation; and here, it was very helpful that you showed all the details, including what you mean by G(z).)

The probability that you want is G(1) - G(-1), but you accidentally replaced G(1) with 1 - G(1), which you had calculated for the previous part of the problem!

The correct calculation is 0.8413 - 0.1587. Right?

You are aboslutely correct. Thank you!

Funfact: In Norway, where I'm from, we use commas for our decimals.
 
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