Math Is Fun Forum

bob's

Member

Registered: 2007-05-31

Posts: 5

help?

Hi this is Bobs again I was wondering if you could check statistics for me! I have asked so many questions thanks for your help! it is jsut that stats is a difficult subject for me!

Data was collected outside a popular new restaurant to determine the mean waiting time to be seated at a table. Assume the data was normally distributed with a mean of 45 min and standard deviation of 12 min
determine the probability that a randomly selected person has to wait less than 20 min
So for this questionI would use:
normalcdf(0,20,45,12)
= 0.0185219157

on to part B

determine the probability that a randomly selected person has to wait more than 1 h?
they have to wait more than an hour
so I cant use the normalCDF thing

thanks once again!

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: help?

Try doing 100% minus normalcdf(0,60,45,12), maybe??
Because if normalcdf is say 55% then 100% - 55% is 45%.
Because the greater and the less of one hour have to add up to 100% of the
probability I think.  Just a guess. I don't do statistics, not for decades.

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igloo myrtilles fourmis

yonski

Member

Registered: 2005-12-14

Posts: 67

Re: help?

We've just been doing this at school, and we always use a table of Z values to calculate this type of problem. Have you done anything with those? If you want me to go through it with that method i can do.

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Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

bob's

Member

Registered: 2007-05-31

Posts: 5

Re: help?

Yes can you please that would be great! I dont really get the Z score!

bob's

Member

Registered: 2007-05-31

Posts: 5

Re: help?

wait! I will give it a try! maybe can you tell me if I did it right?
so:
z= x- mean/ standard deviation
z= (1) - ( 45) / 12
z= -2.75 --> rounded 2.80

so:
corresponding probablitlity is : 0.026

is this right? I hope so! thanks!

yonski

Member

Registered: 2005-12-14

Posts: 67

Re: help?

Nah, you've got the right formula but you've used it wrongly (and put in the the wrong value for X). It's tough to really explain what the Z values represent without using diagrams, so I suggest you ask your teacher or whoever to go over it cos you're fumbling in the dark a bit at the moment (no offence).

Basically this is how you do the first question:

P(X < 20) = P(Z < (20-45)/12)         using the formula you quoted
               = P(Z < -2.08)
               = 1 - P(Z < 2.08)             using the fact the the normal distribution is symmetrical
               = 1 - 0.9812                    look up Z=2.08 in your table and find the corresponding value of theta(Z)
               = 0.0188

and that's your answer

Last edited by yonski (2007-06-07 08:37:56)

---

Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

bob's

Member

Registered: 2007-05-31

Posts: 5

Re: help?

okay can you check:
x--> 1h--> 60 min
mean=45
sd= 12
P(X > 60) = P(Z > (60-45)/12)         
               = P(Z > 1.25)
               = 1 - P(Z < 2.08)             
              Z=1.25 --> 0.90332
but since> you dont need to minus one right?
is this correct? please and thank you! I hope I did it right! dont worry I am going to ask someone

yonski

Member

Registered: 2005-12-14

Posts: 67

Re: help?

No, think about the answer you're giving. How could the probability of them waiting more than 1 hour be 0.90332? This would mean that over 90% of people wait longer than an hour, but you've been told that the mean is 45 minutes!!!

You need to go:

(X > 60) = P(Z > (60-45)/12)         
             = P(Z > 1.25)                           you were right up to this point
             = 1 - P(Z < 1.25)                     
             = 1 - 0.8944
             = 0.1056

Always check that the answer is sensible lol.

---

Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

smileybob

Member

Registered: 2007-06-05

Posts: 2

Re: help?

haha! this is too funny bob's needs help in this and me smileybob need help 2! lol to funny anyways I was wondering if someone could check my work thanks so the question is like so:
The time it takes for a pendulum to complete the one period is simplified by the formula: T=2 pi square root (L/g) Where g is the constant measuring 9.80655m/s^2 and L is the length of the pendulum. The length of the pendulums from a manufacture are normally distributed with a mean of 10cm and a standard deviation of 0.01 cm. Assuming the at a 10cm pendulum gives the correct time, what is the probability that a clock using one of theses pendulums will lose more than 1 minute a day?

so my work so far:

Standard deviation: 0.01cm
Mean= 10cm
T= 2pi [sqrt] L/g
Time increases by: 1/ 24x 60
……
P(x> 10.014)
= p (z> (10.014-10 / 0.01)
= P (Z > 1.4)
Z= 0.919243

But I noticed that you took away 1 from the z right?
Do you think that is necessary here? My next question is this correct?

Best regards
Smiley bob
Thank you!
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smileybob

Member

Registered: 2007-06-05

Posts: 2

Re: help?

hi yonski  can you please check my work? thank you
Best regards
Smiley bob
Thank you!
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My brain is my second favourite organ! what is you first?

misdamike55

Member

Registered: 2007-08-10

Posts: 1

Re: help?

Z Is A Standard Normal Random Variable
How Do You Do This One
P(z>-2.08)