The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 43 o
unces and a standard deviation of 7 ounces. Use the Empirical Rule. Standard Normal Curve with Empirical Values Suggestion: Sketch the distribution in order to answer these questions.
a) 95% of the widget weights lie between and
b) What percentage of the widget weights lie between 12 and 57 ounces? %
c) What percentage of the widget weights lie above 30 ? %
a) 95% of the widget weights lie between and
b) What percentage of the widget weights lie between 12 and 57 ounces? %
c) What percentage of the widget weights lie above 30 ? %
1 answer:
Answer:
a) 95% of the widget weights lie between 29 and 57 ounces.
b) What percentage of the widget weights lie between 12 and 57 ounces? about 97.5%
c) What percentage of the widget weights lie above 30? about 97.5%
Step-by-step explanation:
The empirical rule for a mean of 43 and a standard deviation of 7 is shown below.
a) 29 represents two standard deviations below the mean, and 57 represents two standard deviations above the mean, so, 95% of the widget weights lie between 29 and 57 ounces.
b) 22 represents three standard deviations below the mean, and the percentage of the widget weights below 22 is only 0.15%. We can say that the percentage of widget weights below 12 is about 0. Equivalently we can say that the percentage of widget weights between 12 an 43 is about 50% and the percentage of widget weights between 43 and 57 is 47.5%. Therefore, the percentage of the widget weights that lie between 12 and 57 ounces is about 97.5%
c) The percentage of widget weights that lie above 29 is 47.5% + 50% = 97.5%. We can consider that the percentage of the widget weights that lie above 30 is about 97.5%
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(a) Both conditions are satisfied with <em>x</em> = (1, 0) for and <em>x</em> = (1, 0, 0) for
:
||(1, 0)|| = √(1² + 0²) = 1
max{1, 0} = 1
||(1, 0, 0)|| = √(1² + 0² + 0²) = 1
max{1, 0, 0} = 1
(b) This is the well-known triangle inequality. Equality holds if one of <em>x</em> or <em>y</em> is the zero vector, or if <em>x</em> = <em>y</em>. For example, in , take <em>x</em> = (0, 0) and <em>y</em> = (1, 1). Then
||<em>x</em> + <em>y</em>|| = ||(0, 0) + (1, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2
||<em>x</em>|| + ||<em>y</em>|| = ||(0, 0)|| + ||(1, 1)|| = √(0² + 0²) + √(1² + 1²) = √2
The left side is strictly smaller if both vectors are non-zero and not equal. For example, if <em>x</em> = (1, 0) and <em>y</em> = (0, 1), then
||<em>x</em> + <em>y</em>|| = ||(1, 0) + (0, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2
||<em>x</em>|| + ||<em>y</em>|| = ||(1, 0)|| + ||(0, 1)|| = √(1² + 0²) + √(0² + 1²) = 2
and of course √2 < 2.
Similarly, in you can use <em>x</em> = (0, 0, 0) and <em>y</em> = (1, 1, 1) for the equality, and <em>x</em> = (1, 0, 0) and <em>y</em> = (0, 1, 0) for the inequality.
(c) Recall the dot product identity,
<em>x</em> • <em>y</em> = ||<em>x</em>|| ||<em>y</em>|| cos(<em>θ</em>),
where <em>θ</em> is the angle between the vectors <em>x</em> and <em>y</em>. Both sides are scalar, so taking the norm gives
||<em>x</em> • <em>y</em>|| = ||(||<em>x</em>|| ||<em>y</em>|| cos(<em>θ</em>)|| = ||<em>x</em>|| ||<em>y</em>|| |cos(<em>θ</em>)|
Suppose <em>x</em> = (0, 0) and <em>y</em> = (1, 1). Then
||<em>x</em> • <em>y</em>|| = |(0, 0) • (1, 1)| = 0
||<em>x</em>|| • ||<em>y</em>|| = ||(0, 0)|| • ||(1, 1)|| = 0 • √2 = 0
For the inequality, recall that cos(<em>θ</em>) is bounded between -1 and 1, so 0 ≤ |cos(<em>θ</em>)| ≤ 1, with |cos(<em>θ</em>)| = 0 if <em>x</em> and <em>y</em> are perpendicular to one another, and |cos(<em>θ</em>)| = 1 if <em>x</em> and <em>y</em> are (anti-)parallel. You get everything in between for any acute angle <em>θ</em>. So take <em>x</em> = (1, 0) and <em>y</em> = (1, 1). Then
||<em>x</em> • <em>y</em>|| = |(1, 0) • (1, 1)| = |1| = 1
||<em>x</em>|| • ||<em>y</em>|| = ||(1, 0)|| • ||(1, 1)|| = 1 • √2 = √2
In , you can use the vectors <em>x</em> = (1, 0, 0) and <em>y</em> = (1, 1, 1).