A college writing seminar increased its size by 50\PP, percent from the first to the second day. If the total number of students
1 answer:
Answer: There were 10 students in the class on the first day.
Step-by-step explanation:
Let x be the number of students of the first day.
Given: A college writing seminar increased its size by 50 percent from the first to the second day.
i.e. Number of students on second day = (Number of students on first day)+(50% of Number of students on first day)
= x +50% of x
= x+0.50x
= (1.50)x
=1.50x
Since, it is given that the total number of students in the seminar on the second day was 15.
i.e.
Hence, there were 10 students in the class on the first day.
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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.
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
- The population standard deviation is
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
For the given parameters, we have:
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:
(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
See the graph below. The shaded area is the region that represents the proportion asked in the question.
