# determining mean without Z score

#### Ade

##### New Member
Hello,
Not looking for the answer, I've just got myself confused. The scenario I have is: I have been given the Standard deviation figure 14.62 (years), normal distribution, n=26, 32% of 26 (n) are aged less than 50. What is the mean? (i.e I don't have z or the mean)

I've initially tried to 'back engineer' a standard calculation example I had to find the proportion of n below 50 with a mean of 53.4, Standard Deviation 14.2, but it didn't seem on track. Then tried to make sence of calculating z with Tables of the Normal Distribution, but I'm missing something, and just can't get me head around it now.

Regards,
Ade

#### JohnM

##### TS Contributor
The probability of getting a score less than 50 (call it x) is 0.32.

z(0.32) = -0.468

z = (x - mu) / s

You now have z, s, and x. Solve for mu.

#### Ade

##### New Member
Thanks John, I knew it had to be something simple, and I'm stilkl doing something wrong as I keep getting an answer of 6.55 as the mean and that's with two different formula's, the one you provided and another I found in a text. But the 6.55 can't be right as a mean age if the standard deviation is 14.62 as that would make the lower age less than 0. I think I've spent too much time looking at statistics and got a little frezzled? What am I missing?
Regards,
Ade

#### JohnM

##### TS Contributor
z = (x - mu) / s

z = -0.468
x = 50
s = 14.62

-0.468 = (50 - mu) / 14.62

(14.62)(-0.468) = 50 - mu

mu = 50 - (14.62)(-0.468)
= 50 - (-6.842)
= 50 + 6.842
= 56.842

#### Ade

##### New Member
Thankyou. I am looking too hard, think I'll take a break from study.