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.