Diana made a recipe that yields 6 and 1 over 2 cups. If each serving is 1 over 4 cup, which expression will help Diana determine
2 answers:
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Answer:
General Formulas and Concepts:
<u>Pre-Alg</u>
- Order of Operations: BPEMDAS
<u>Alg I</u>
- Slope Formula:
Step-by-step explanation:
<u>Step 1: Define</u>
Point (2, 6)
Point (1, 9)
<u>Step 2: Find slope </u><em><u>m</u></em>
- Substitute:
- Subtract:
- Divide:
Answer:
The value <em>155</em> is zero standard deviations from the [population] mean, because , and therefore
.
Step-by-step explanation:
The key concept we need to manage here is the z-scores (or standardized values), and we can obtain a z-score using the next formula:
[1]
Where
- z is the <em>z-score</em>.
- x is the <em>raw score</em>: an observation from the normally distributed data that we want <em>standardize</em> using [1].
is the <em>population mean</em>.
is the <em>population standard deviation</em>.
Carefully looking at [1], we can interpret it as <em>the distance from the mean of a raw value in standard deviations units. </em>When the z-score is <em>negative </em>indicates that the raw score, <em>x</em>, is <em>below</em> the population mean, . Conversely, a <em>positive</em> z-score is telling us that <em>x</em> is <em>above</em> the population mean. A z-score is also fundamental when determining probabilities using the <em>standard normal distribution</em>.
For example, think about a z-score = 1. In this case, the raw score is, after being standardized using [1], <em>one standard deviation above</em> from the population mean. A z-score = -1 is also one standard deviation from the mean but <em>below</em> it.
These standardized values have always the same probability in the <em>standard normal distribution</em>, and this is the advantage of using it for calculating probabilities for normally distributed data.
A subject earns a score of 155. How many standard deviations from the mean is the value 155?
From the question, we know that:
- x = 155.
.
.
Having into account all the previous information, we can say that the raw score, <em>x = 155</em>, is <u><em>zero standard deviations units from the mean.</em></u> <u><em>The subject earned a score that equals the population mean.</em></u> Then, using [1]:
As we say before, the z-score "tells us" the distance from the population mean, and in this case this value equals zero:
Therefore
So, the value 155 is zero standard deviations <em>from the [population] mean</em>.