Answered

Nicole is in her first semester at a certain university and is taking a calculus course with a large enrollment. She just took the first midterm exam and is nervous about her score. Among all of the students in the course, the mean of the exam was 77 with a standard deviation of 9. Nicole scored a 70 on this exam.

(a) Find the [tex]$z$[/tex]-score of Nicole's exam score relative to the exam scores among all the students in the course. Round your answer to two decimal places.
[tex]\[ z = \square \][/tex]

(b) Fill in the blanks to interpret the [tex]$z$[/tex]-score of Nicole's exam score. Make sure to express your answer in terms of a positive number of standard deviations.

Nicole's exam score was [tex]$\square$[/tex] standard deviations (Choose one: above/below) the mean exam score among all students in the course.



Answer :

GinnyAnswer
Let's solve the problem step-by-step.

### Part (a): Finding the [tex]$z$[/tex]-score

Given the information:
- Mean score ([tex]\(\mu\)[/tex]) = 77
- Standard deviation ([tex]\(\sigma\)[/tex]) = 9
- Nicole's score ([tex]\(X\)[/tex]) = 70

The formula to calculate the [tex]$z$[/tex]-score is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]

Substituting the values:
[tex]\[ z = \frac{70 - 77}{9} \][/tex]
[tex]\[ z = \frac{-7}{9} \][/tex]
[tex]\[ z = -0.78 \][/tex]

So the [tex]$z$[/tex]-score of Nicole's exam score is:
[tex]\[ z = -0.78 \][/tex]

### Part (b): Interpreting the [tex]$z$[/tex]-score

A [tex]$z$[/tex]-score tells us how many standard deviations a particular score is from the mean. The [tex]$z$[/tex]-score in this case is [tex]\(-0.78\)[/tex], which indicates that Nicole's score is below the mean. To express this in terms of a positive number of standard deviations:

- Take the absolute value of the [tex]$z$[/tex]-score: [tex]\(|-0.78| = 0.78\)[/tex]
- Nicole's score is 0.78 standard deviations below the mean.

The interpretation will be:
Nicole's exam score was [tex]\(0.78\)[/tex] standard deviations [tex]\(\text{below}\)[/tex] the mean exam score among all students in the course.

Other Questions