# Thread: normal distribution

1. ## normal distribution

Hi all,

I've solved the question but I'm not sure whether or not my answer is correct. I need your help.

The length of time it takes to be seated at a local restaurant on Friday night is normally distributed with a mean of 15 minutes and a standard deviation of 4.75 minutes.

a. What is the probability that you have to wait more than 20 minutes to be seated?
b. What is the probability that you have to wait between 13 and 16 minutes to be seated?

a)

x= length of time

μ= 15
σ= 4.75
P(x>20)

For x=20, z=(20-15)/4.75 -> 1.052
P(x>15)= P(z> 1.052) = [total area] - [area to the left of z = 1]
1- 0.8536 = 0.1464

b)
x= length of time
μ = 15
σ= 4.75

P(13<x<16) For x=13, z= (13-15)/4.75 -> -0.421
For x=16, z= 16-15/4.75 -> 0.210
P(13<x16) = P(-0.412<z< 0.210)
0.416-0.340= 0.076

2. ## Re: normal distribution

Originally Posted by Liliann
Hi all,

I've solved the question but I'm not sure whether or not my answer is correct. I need your help.

The length of time it takes to be seated at a local restaurant on Friday night is normally distributed with a mean of 15 minutes and a standard deviation of 4.75 minutes.

a. What is the probability that you have to wait more than 20 minutes to be seated?
b. What is the probability that you have to wait between 13 and 16 minutes to be seated?

a)

x= length of time

μ= 15
σ= 4.75
P(x>20)

For x=20, z=(20-15)/4.75 -> 1.052
P(x>15)= P(z> 1.052) = [total area] - [area to the left of z = 1]
1- 0.8536 = 0.1464

This is correct

b)
x= length of time
μ = 15
σ= 4.75

P(13<x<16) For x=13, z= (13-15)/4.75 -> -0.421
For x=16, z= 16-15/4.75 -> 0.210
P(13<x16) = P(-0.412<z< 0.210)
0.416-0.340= 0.076

This is incorrect. You correctly computed the z-scores but did not convert those into probabilities correctly.

3. ## Re: normal distribution

Originally Posted by romsek
[/FONT][/SIZE][/CENTER][/CENTER]

This is correct

This is incorrect. You correctly computed the z-scores but did not convert those into probabilities correctly.
Thanks a lot.

0.416+0.340 = 0.756 is correct?

4. ## Re: normal distribution

Originally Posted by Liliann
Thanks a lot.

0.416+0.340 = 0.756 is correct?
why would you add?

The numbers I get are

$pr[16]\approx 0.583372$

$pr[13]\approx 0.336858$ you seem to have gotten this value correct

the probability you are after is the difference of these two numbers

$pr[16]-pr[13]\approx 0.246513$

5. ## Re: normal distribution

Originally Posted by romsek
why would you add?

The numbers I get are

$pr[16]\approx 0.583372$

$pr[13]\approx 0.336858$ you seem to have gotten this value correct

the probability you are after is the difference of these two numbers

$pr[16]-pr[13]\approx 0.246513$
Thanks again.

6. ## Re: normal distribution

Pr [X>20]=Pr[N(0,1)>20-15/4.75]=Pr[N(0,1)>1.0526]=1-.8537=.1463
Pr[X<16]-Pr[X<13]=Pr[N(0,1)<16-15/4.75]-Pr[N(0,1)<13-15/4.75]=
Pr[N(0,1)<.2105]-Pr[N(0,1)<-.4211]=.5834-.3368=.2466

7. ## Re: normal distribution

Originally Posted by RShnay
Pr [X>20]=Pr[N(0,1)>20-15/4.75]=Pr[N(0,1)>1.0526]=1-.8537=.1463
Pr[X<16]-Pr[X<13]=Pr[N(0,1)<16-15/4.75]-Pr[N(0,1)<13-15/4.75]=
Pr[N(0,1)<.2105]-Pr[N(0,1)<-.4211]=.5834-.3368=.2466
Thank you very much.