The mean of 12 teachers in a school is 38 years with a standard deviation of 5.3. Ms. Data is 24 years old and Mr. Management is 53 years of age. Calculate the z-score of Ms. Data and the percentage of teachers that Mr. Management is older than if the teacher's age follows a normal distribution.
I'm stuck on this question. If someone could explain the solution in plain English that'll be great!
Thanks!
Right, so the "z-score" is, as @JMoravitz noted, the "number of standard deviations above or below the mean." They give you, in the problem what 1 standard deviation is.
The formula, also given in the comment, is $\frac{x-\mu}{\sigma}=z$. We are taking the difference between $x$, here the age of Ms. Data, and the mean, and finding out how many standard deviations fit into the difference.
In general $x$ is a single datapoint/observation. We want to find out how many standard deviation's (how many 5.3's) fit in between the mean, and the observation.
Now, as to the second part of the question, you also need to find the $z$-score for Mr. Management. Why? Because, and I hope they talked about this in class, in order to find the percentages of teachers whose age is less than Mr. Management, you need to use a table or calculator (or integral).
The percentage of teachers less than Mr. Management's age is the same thing as saying, the probability that a teacher at the school is younger than Mr. Management. In statistics formalism, we might say, $P(A<M)$. We are finding the following probability:
The only difference, is that Mr. Management is on the right side of the bump, because he is older than the mean age. We can find his z-score, then find the corresponding left-tailed probability (the area to the left).
You see, the standard normal distribution is defined as being centered around a mean of $0$, with a standard deviation of $1$. The $z$-score of Mr. Management's age, is equivalent to his position on the standard normal distribution.