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What is the square root of 1/121

icang [17]

Answer:

11

Step-by-step explanation:

example 11×11 = 121

therefore squreroots = 11

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3 years ago

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The amount of protein that an individual must consume is different for every person. There are solid theoretical ideas that sugg

amid [387]

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.

$\\ 0.239 =\frac{239}{1000}\;or\;23.9\%$

Step-by-step explanation:

From the question, we have the following information:

The distribution for protein requirement is normally distributed.
The population mean for protein requirement for adults is $\\ \mu= 0.65 gP*kg^{-1}*d^{-1}$
The population standard deviation is $\\ \sigma =0.07 gP*kg^{-1}*d^{-1}$

We have here that protein requirements in adults is normally distributed with defined parameters. The question is about the proportion of the population that has a requirement less than $\\ x = 0.60 gP*kg^{-1}*d^{-1}$ .

For answering this, we need to calculate a z-score to obtain the probability of the value x in this distribution using a standard normal table available on the Internet or on any statistics book.

<h3>z-score</h3>

A z-score is expressed as

$\\ z = \frac{x - \mu}{\sigma}$

For the given parameters, we have:

$\\ z = \frac{0.60 - 0.65}{0.07}$

$\\ z = -0.7142857$

<h3>Determining the probability</h3>

With this value for z 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 symmetrical.

Tables have values for z with two decimal places, then, for z = 0.7142857, we need to rewrite it as z = 0.71. For this value, the standard normal table 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:

$\\ P(x$