# I am not sure how to solve for a probability when N is not given (beginner help).

#### annaml

##### New Member
I am having the most trouble with part "a)" but I am comparing the two answers so if you feel inclined to discuss part "b)" as well, I would certainly appreciate it!

The problem:
"The average annual age of doctors in a certain hospital is 43.0 years old with a standard deviation of 3.0 years. Assume that the variable is normally distributed.
(a) Find the probability that the age of a randomly selected doctor is less than 42.32 years.
(b) Find the probability that the mean age of a sample of 25 randomly selected doctors is less than 42.32 years."

I know that "normal distribution" entails that mean=median=mode which I'm thinking might help me solve for N but I'm still stumped. Thank you!

#### jamesmartinn

##### Member
From your question, I think you can take the supplied values as population parameters.

The "part a" is then a straight forward calculation using a Z-score and a table of values from a standard normal distribution.

#### Dragan

##### Super Moderator
Hint: Additionally, think of the sample size in part (a) as N = 1 and part (b) N = 25.

#### annaml

##### New Member
Oh ok, I get what you mean. So Z= (value-mean)/SD --> Z= (42.32-43.0)/3.0. Then find value on Z-score table to get probability.

#### Dragan

##### Super Moderator
Oh ok, I get what you mean. So Z= (value-mean)/SD --> Z= (42.32-43.0)/3.0. Then find value on Z-score table to get probability.
Well, no...you're not there quite yet: You want Z= (42.32-43.0)/(3.0/Sqrt[1]).....so you can generalize.

#### annaml

##### New Member
Ah yes! I see it now, I was looking at an example in the text where they dropped the sqrt(1). Thank you!!

#### Dragan

##### Super Moderator
Ah yes! I see it now, I was looking at an example in the text where they dropped the sqrt(1). Thank you!!
Okee Dokee.