If the sin 30° is 1 over 2, then the cos ____° = _____. (4 points) 60° ; 1 over 2 30° ; square root 2 over 2 60° ; square root 3
2 answers:
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Answer:
The "probability that a given score is less than negative 0.84" is .
Step-by-step explanation:
From the question, we have:
- The random variable is <em>normally distributed</em> according to a <em>standard normal distribution</em>, that is, a normal distribution with
and
.
- We are provided with a <em>z-score</em> of -0.84 or
.
Preliminaries
A z-score is a standardized value, i.e., one that we can obtain using the next formula:
[1]
Where
- <em>x</em> is the <em>raw value</em> coming from a normal distribution that we want to standardize.
- And we already know that
and
are the mean and the standard deviation, respectively, of the <em>normal distribution</em>.
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:
- Find the <em>cumulative standard normal table.</em>
- In the first column of the table, use -0.8 as an entry.
- Then, using the first row of the table, find -0.04 (which determines the second decimal place for the z-score.)
- The intersection of these two numbers "gives us" the cumulative probability for z or
.
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
.
