3.) In a recent contest, the mean score was 210 and the standard deviation was 25.

A.) Find the Z score of john who scored 190.

In order to find the Z score, we use the Z formula that we learned in class:

Z equals X minus mu (mu is the same as the mean) divided by standard deviation.

We are given a mean score of 210 and a standard deviation of 25 in number 3.

In A we are given our X which is 190. Our X represents John’s score.

We plug the numbers in and solve: 190-210/25=-0.8     —-> remember always follow PEMDAS (parenthesis, exponent, multiplication, division, addition and subtraction.) We subtract 190 -210 first and then divide by 25.

B.) Find the Z score of Bill who scored 270.

Again we are asked to find the Z score. We follow the same formula used in 3a.

Z equals X minus mu divided by standard deviation.

Here our mean and standard deviation are the same but our X is different.

Our mean (which is given) is 210 and our standard deviation (also given) is 25

Our X was Bill’s score which was 270.

Now we plug our numbers into the formula:

270-210/25=2.4

Here PEMDAS should also be used which means we subtract before we divide.

C.) If Mary had a Z-score of 1.25, what was Mary’s score.

Here unlike A and B, we are given Mary’s Z score and are asked to find her X.

In this case, we use the X formula:

X equals mu plus Z times standard deviation

Z=1.25

mu=210 (given in original problem)

Standard Deviation=25 (also given in original problem)

When we plug the numbers is we get:

210+1.25(25)=241.25

*Here it is important to do PEMDAS as well. First we multiply 1.25(25) and get 31.25 and then we add 210 and get a final answer of 241.25

*I had to write out the formulas because there is no mu symbol nor is there a standard deviation symbol on the computer so here is a picture of the work!

Hope this helped 🙂