Given:
There are given that the parent functions as a cosine function:
Where,
The amplitude of the function is 9.
The vertical shift is 11 units down.
Explanation:
To find the cosine function, we need to see the standard form of the cosine function:
Where,
a is the amplitude of the function,
Now,
According to the question:
The amplitude of the function is 9, which means:
The vertical shift is 11 units down, which means:
For period:
Final answer:
Hence, the cosine function is shown below;
Answer:
The "probability that a given score is less than negative 0.84" is .
Step-by-step explanation:
From the question, we have:
Preliminaries
A z-score is a standardized value, i.e., one that we can obtain using the next formula:
[1]
Where
A <em>z-score</em> represents the <em>distance</em> from in <em>standard deviations</em> units. When the value for z is <em>negative</em>, it "tells us" that the raw score is <em>below</em>
. Conversely, when the z-score is <em>positive</em>, the standardized raw score, <em>x</em>, is <em>above</em> the mean,
.
Solving the question
We already know that or that the standardized value for a raw score, <em>x</em>, is <em>below</em>
in <em>0.84 standard deviations</em>.
The values for probabilities of the <em>standard normal distribution</em> are tabulated in the <em>standard normal table, </em>which is available in Statistics books or on the Internet and is generally in <em>cumulative probabilities</em> from <em>negative infinity</em>, - , to the z-score of interest.
Well, to solve the question, we need to consult the <em>standard normal table </em>for . For this:
Therefore, we obtain for this z-score, or a slightly more than 20% (20.045%) for the "probability that a given score is less than negative 0.84".
This represent the area under the <em>standard normal distribution</em>, , at the <em>left</em> of <em>z = -0.84</em>.
To "draw a sketch of the region", we need to draw a normal distribution <em>(symmetrical bell-shaped distribution)</em>, with mean that equals 0 at the middle of the distribution, , and a standard deviation that equals 1,
.
Then, divide the abscissas axis (horizontal axis) into <em>equal parts</em> of <em>one standard deviation</em> from the mean to the left (negative z-scores), and from the mean to the right (positive z-scores).
Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, , from it). All the area to the left of this value must be shaded because it represents
and that is it.
The below graph shows the shaded area (in blue) for for
.
