Rectangle A measures 8 inches by 4 inches. Rectangle B is a scaled copy of Rectangle A. Select all of the measurement pairs that
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The answer assumes that the question is about the <em>normal distribution</em>.
Answer:
96% of the data in any data set <em>normally distributed</em> is 2.05 <em>standard deviations</em> <em>above</em> and <em>below</em> the mean.
Step-by-step explanation:
The key to solving this question is having into account that exists the <em>standard normal distribution</em> that permits obtaining any probability from any normally distributed data, doing the following transformation:
That is, in this case, we subtract a given value <em>x</em> from the <em>population mean</em> and then divide the result by the <em>population standard deviation</em>. This is a z-score, and this value is associated with the probability for a <em>standard normal distribution</em>, with a population mean = 0 and population standard deviation = 1.
Fortunately, for the <em>standard normal distribution</em>, there is associated a ubiquitous <em>standard normal table</em> for possible values of <em>z</em> and the corresponding probability. Then, knowing that the data are <em>normally distributed</em> and having both distribution <em>parameters</em>, namely, the <em>population mean</em> and <em>population standard deviation</em>, we can consult any standard normal table to find the probabilities for any data distributed following the normal distribution.
However, even without previously know the values of the normal parameters, the standard normal distribution can tell us how many standard deviations from the mean are 96.00% of the data for every normally distributed population.
<h3>Solving the question</h3>Having all this information at hand, we know that <em>the z-score value tells us how many standard deviations</em> <em>from the mean</em> are the data normally distributed. As a result, we can determine how many standard deviations below and above the population mean represent the 96.00% of the cases for this normal distribution consulting a <em>cumulative standard normal table from the mean</em>.
If we divide 96.00/2 = 48, that is, 0.48, we need to find the z-score for this probability consulting a <em>cumulative standard normal table from the mean</em>. The values for a z-score for that probability is z=2.05, approximately. So, since the normal distribution is also symmetrical, those values are above and below the mean, that is, z =2.05 (above) and z=-2.05(below).
Thus, 96% of the data in any data set normally distributed is 2.05 <em>standard deviations</em> <em>above</em> and <em>below</em> the mean.
(a) See the attached sketch. Each shell will have a radius <em>y</em> chosen from the interval [2, 4], a height of <em>x</em> = 2/<em>y</em>, and thickness ∆<em>y</em>. For infinitely many shells, we have ∆<em>y</em> converging to 0, and each super-thin shell contributes an infinitesimal volume of
2<em>π</em> (radius)² (height) = 4<em>πy</em>
Then the volume of the solid is obtained by integrating over [2, 4]:
(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - <em>x</em> (this is the distance between a given <em>x</em> value in the orange shaded region to the axis of revolution) and a height of 8 - <em>x</em> ³ (and this is the distance between the line <em>y</em> = 8 and the curve <em>y</em> = <em>x</em> ³). Then each shell has a volume of
2<em>π</em> (9 - <em>x</em>)² (8 - <em>x</em> ³) = 2<em>π</em> (648 - 144<em>x</em> + 8<em>x</em> ² - 81<em>x</em> ³ + 18<em>x</em> ⁴ - <em>x</em> ⁵)
so that the overall volume of the solid would be
I leave the details of integrating to you.
