Question on z-score value in probability I am given that a student counts the number of motor vehicles passing in the street and he gets following values for the following 10 intervals (1 interval = 1 min):

0-1= 25

1-2= 27

2-3= 28

3-4= 20

4-5 = 22

5-6 = 20

6-7= 29

7-8= 28

8-9=22

9-10=23

Now, the question asks me to calculate the probability of getting a mean greater than 2 standard deviations of the mean from this count if the student were to take another set of 10 measurements. 

My approach: 
I calculated the sample mean from above and got 24.4 cars/min.
The sample standard deviation was 3.44 cars/min.
Now, my new mean  is supposed to be 31.28 cars/min
Then, I tried to use z score and used the formula: 

z = (New mean-Old mean)/(Sample deviation/sqrt(20)), but doing so, my z score is 8.94, and I do not know what it means cause it says my probability is almost 0. What is wrong in my approach?
You should have divided the sample deviation by $\sqrt{10}$, not $\sqrt{20}$.

Regardless, the $z$-score is VERY high ($6.32)$ and you will compute a probability close to $0$.

This means that it is SUPER unlikely.
You should have divided the sample deviation by $\sqrt{10}$, not $\sqrt{20}$.

Regardless, the $z$-score is VERY high ($6.32)$ and you will compute a probability close to $0$.

This means that it is SUPER unlikely.