Emily wants to watch her intake of fat per day.
2 answers:
<u>Answer:</u>
68.5%
<u>Step-by-step explanation:</u>
We are given that Emily estimates that her dinner will have 48 grams of fat.
From the given guideline of the recommended daily allowances, we can see that the recommended allowance of fat is 70 grams.
We are to find the percentage of her total daily allowance of fat will the dinner be.
= 68.5%
Answer:
The answer is 69%.
Step-by-step explanation:
Emily wants to watch her intake of fat per day.
She estimates that the dinner will have 48 grams of fat.
As per the guidelines given, the fat intake of women can be 70 grams.
Emily will be taking 48 grams of 70 grams.
So, the percentage of her total daily allowance of fat that she will consume for dinner will be =
And rounding this to whole number we get 69%.
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Answer:
The students have to score 0.74 standard deviations above the mean to be publicly recognized.
Step-by-step explanation:
A random variable <em>X</em> is said to have a normal distribution with parameters <em>µ</em> (mean) and <em>σ</em>² (variance).
If , then
, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is,
.
The distribution of these <em>z</em>-scores is known as the standard normal distribution.
The <em>z</em>-score is a standardized form of the raw score, <em>X</em>. It is a numerical measurement of the relationship between a value (<em>X</em>) and the mean (<em>µ</em>) in terms of the standard deviation (<em>σ</em>). A <em>z</em>-score of -1 implies that the data value is 1 standard deviation below the mean. And a <em>z</em>-score of 1 implies that the data value is 1 standard deviation above the mean.
Let <em>X</em> be defined as the scores of students at the National Financial Capability Challenge Exam.
It is provided that the students who score in the top 23% are recognized publicly for their achievement by the Department of the Treasury.
That is, P (X > x) = 0.23.
⇒
⇒
The value of <em>z</em> for this probability value is:
<em>z</em> = 0.74.
*Use a <em>z</em>-table.
Thus, the students have to score 0.74 standard deviations above the mean to be publicly recognized.
