Statistics Normal Distribution.

Question

The average age of a passenger on the Euro rail passenger train is 29.4 years. If the distribution of ages is
normal and 20% of the passengers are older than 33.6 years old, find the standard deviation.

 

 

I have no clue as to what I should start or try to seek.  As I know the 20% of travelers are more then 33.6 , I can assume that x random vairable for 20%  should be 33.7. The z score for the pro seems to be -80.413. I do not understand how .

I think it is 1-0.8143 and I will have my z score of that point which is the higher part of the date 20%.

 

I do not know where to go from here. Any help is appreciated.

Answer 1

Syed,

 

You're on the right track to use the z-score to solve this problem, but you got lost in the details.

 

Let A = Euro rail passenger age distribution given as the Normal distribution N(μ=29.4, σ), whose mean (μ) is specified, but whose standard deviation (σ) must be computed from other data to follow.

Let Z = (A - μ)/σ be the corresponding standardized Z-score distributed N(0, 1) for which Tabular Data are available.

Let A* = 33.6 = age that is only exceeded by 20 % of all the passengers.

 

Thus,

Z* = (A* - μ)/σ  =  (33.6 - 29.4)/σ ,  so Pr[Z > Z*] = 0.20 , or Pr[0≤ Z ≤ Z*] = 0.30 given the Z-Normal distribution

is symmetrical about zero and we only need to work with the right side where Z* lies.  

The left side already satisfies Z ≤ Z*. 

 

In the probability table for the Z-standardized Normal, interpolated value Z* = 0.8415 satisfies the condition that 80% of its area is below Z* and 20% above Z*.

 

So the unknown σ = (A* -  μ)/Z* = (33.6 - 29.4)/0.8415 = 4.2/0.8415 = 4.99  or call it 5.0

 

BTW, for A*, I used 33.6, you used 33.7. This is an arbitrary choice. But someone could be 1 day older than 33.6 (33 and 219 days), or 1 hour, etc. So the 33.6 choice is somewhat "less arbitrary" than 33.7.

Answer 2

Looking at what we know, 20% of passengers are older than 33.6. Now, since 33.6-29.4=4.2 years and since 29.4-4.2=25.2, then 20% of our passengers will be less than 25.2 years. So we know the bottom 20% and the top 20% and are looking to determine something about the 60% in the middle. Looking for a z-score that corresponds with an Area of 60%, we see that z=0.85 is 60.47% which is close, but z=0.845 might be even closer. That means that 4.2 deviation from the mean corresponds to a z-score of z=0.85. In order to find the standard deviation, we want to find a z-score of z=1. The general formula for z-score is z=(observed-expected)/SE or SE=(observed-expected)/z Solving for SE=(33.6-29.4)/0.845 => SE=4.97 So, we can now figure out what age that a z-score of z=1 (aka 1 standard deviation) would correspond to! Using z=(observed-expected)/SE and leaving the "observed" as a variable, we have 1=(observed-29.4)/4.97 or 4.97=(observed-29.4) or 34.37=observed. Thus, someone who is 34.37 years has a z=1, and is thus 1 standard deviation away. But the value of the standard deviation is 34.37-29.4 or 4.97 years.