Answer:
The proportion of the population that have a protein requirement less than 0.60 g P • kg-1 • d-1 is 0.239, that is, 239 persons for every 1000, or simply 23.9% of them.
![\\ 0.239 =\frac{239}{1000}\;or\;23.9\%](https://tex.z-dn.net/?f=%20%5C%5C%200.239%20%3D%5Cfrac%7B239%7D%7B1000%7D%5C%3Bor%5C%3B23.9%5C%25)
Step-by-step explanation:
From the question, we have the following information:
- The distribution for protein requirement is <em>normally distributed</em>.
- The population mean for protein requirement for adults is
![\\ \mu= 0.65 gP*kg^{-1}*d^{-1}](https://tex.z-dn.net/?f=%20%5C%5C%20%5Cmu%3D%200.65%20gP%2Akg%5E%7B-1%7D%2Ad%5E%7B-1%7D)
- The population standard deviation is
![\\ \sigma =0.07 gP*kg^{-1}*d^{-1}](https://tex.z-dn.net/?f=%20%5C%5C%20%5Csigma%20%3D0.07%20gP%2Akg%5E%7B-1%7D%2Ad%5E%7B-1%7D)
We have here that protein requirements in adults is normally distributed with defined parameters. The question is about <em>the proportion</em> <em>of the population</em> that has a requirement less than
.
For answering this, we need to calculate a <em>z-score</em> to obtain the probability of the value <em>x </em>in this distribution using a <em>standard normal table</em> available on the Internet or on any statistics book.
<h3>z-score</h3>
A z-score is expressed as
![\\ z = \frac{x - \mu}{\sigma}](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%20%5Cfrac%7Bx%20-%20%5Cmu%7D%7B%5Csigma%7D)
For the given parameters, we have:
![\\ z = \frac{0.60 - 0.65}{0.07}](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%20%5Cfrac%7B0.60%20-%200.65%7D%7B0.07%7D)
![\\ z = \frac{0.60 - 0.65}{0.07}](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%20%5Cfrac%7B0.60%20-%200.65%7D%7B0.07%7D)
![\\ z = -0.7142857](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%20-0.7142857)
<h3>Determining the probability</h3>
With this value for <em>z</em> at hand, we need to consult a standard normal table to determine what the probability of this value is.
The value for z = -0.7142857 is telling us that the requirement for protein is below the population mean (negative sign indicates this). However, most standard normal tables give a probability that a statistic is less than z and for values greater than the mean (in other words, positive values). To overcome this, we need to take the complement of the probability given for z-score z = 0.7142857, that is, subtract from 1 this probability, which is possible because the normal distribution is <em>symmetrical</em>.
Tables have values for <em>z</em> with two decimal places, then, for z = 0.7142857, we need to rewrite it as z = 0.71. For this value, the <em>standard normal table</em> gives a value of P(z<0.71) = 0.76115.
Therefore, the cumulative probability for values less than x = 0.60 which corresponds to a z-score = -0.7142857 is approximately:
![\\ P(x](https://tex.z-dn.net/?f=%20%5C%5C%20P%28x%3C0.60%29%20%3D%201%20-P%28z%3C0.71%29%20%3D%201%20-%200.76115%20%3D%200.23885)
(rounding to three decimal places)
That is, the proportion of the population that have a protein requirement less than 0.60 g P • kg-1 • d-1 is
![\\ 0.239 =\frac{239}{1000}\;or\;23.9\%](https://tex.z-dn.net/?f=%20%5C%5C%200.239%20%3D%5Cfrac%7B239%7D%7B1000%7D%5C%3Bor%5C%3B23.9%5C%25)
See the graph below. The shaded area is the region that represents the proportion asked in the question.