Answer:
don't have answer but have how to do them
Step-by-step explanation:
What are z-scores?
A z-score measures exactly how many standard deviations above or below the mean a data point is.
Here's the formula for calculating a z-score:
z=\dfrac{\text{data point}-\text{mean}}{\text{standard deviation}}z=
standard deviation
data point−mean
z, equals, start fraction, start text, d, a, t, a, space, p, o, i, n, t, end text, minus, start text, m, e, a, n, end text, divided by, start text, s, t, a, n, d, a, r, d, space, d, e, v, i, a, t, i, o, n, end text, end fraction
Here's the same formula written with symbols:
z=\dfrac{x-\mu}{\sigma}z=
σ
x−μ
z, equals, start fraction, x, minus, mu, divided by, sigma, end fraction
Here are some important facts about z-scores:
A positive z-score says the data point is above average.
A negative z-score says the data point is below average.
A z-score close to 000 says the data point is close to average.
A data point can be considered unusual if its z-score is above 333 or below -3−3minus, 3. [Really?]
Want to learn more about z-scores? Check out this video.
Example 1
The grades on a history midterm at Almond have a mean of \mu = 85μ=85mu, equals, 85 and a standard deviation of \sigma = 2σ=2sigma, equals, 2.
Michael scored 868686 on the exam.
Find the z-score for Michael's exam grade.
\begin{aligned}z&=\dfrac{\text{his grade}-\text{mean grade}}{\text{standard deviation}}\\ \\ z&=\dfrac{86-85}{2}\\ \\ z&=\dfrac{1}{2}=0.5\end{aligned}
z
z
z
=
standard deviation
his grade−mean grade
=
2
86−85
=
2
1
=0.5
Michael's z-score is 0.50.50, point, 5. His grade was half of a standard deviation above the mean.
Example 2
The grades on a geometry midterm at Almond have a mean of \mu = 82μ=82mu, equals, 82 and a standard deviation of \sigma = 4σ=4sigma, equals, 4.
Michael scored 747474 on the exam.
Find the z-score for Michael's exam grade.
\begin{aligned}z&=\dfrac{\text{his grade}-\text{mean grade}}{\text{standard deviation}}\\ \\ z&=\dfrac{74-82}{4}\\ \\ z&=\dfrac{-8}{4}=-2\end{aligned}
z
z
z
=
standard deviation
his grade−mean grade
=
4
74−82
=
4
−8
=−2
Michael's z-score is -2−2minus, 2. His grade was two standard deviations below the mean.
https://www.khanacademy.org/math/statistics-probability/modeling-distributions-of-data/z-scores/a/z-scores-review
this should help
