The data set below shows the number of books checked out from a library during the first two weeks of the month: 10, 89, 80, 95,
2 answers:
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The number of pages in the book is 180
<u>Explanation:</u>
The first thing we should do is to convert these fractions to percentage
3/4 of the remainder on thursday:
Thursday: (100 - (13.33 + 33.33 + 22.22) * (3/4) = 23.33%
he still has 14 pages left to read on friday
Friday: [100 - (13.33 + 33.33 + 22.22) ] - 23.33 = 7.77%
Which means that 14 pages represent 7.77% of the book.
Therefore the number of pages of the book is given by:
Therefore, number of pages in the book is 180
Answer:
The students have to score 0.74 standard deviations above the mean to be publicly recognized.
Step-by-step explanation:
A random variable <em>X</em> is said to have a normal distribution with parameters <em>µ</em> (mean) and <em>σ</em>² (variance).
If , then
, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is,
.
The distribution of these <em>z</em>-scores is known as the standard normal distribution.
The <em>z</em>-score is a standardized form of the raw score, <em>X</em>. It is a numerical measurement of the relationship between a value (<em>X</em>) and the mean (<em>µ</em>) in terms of the standard deviation (<em>σ</em>). A <em>z</em>-score of -1 implies that the data value is 1 standard deviation below the mean. And a <em>z</em>-score of 1 implies that the data value is 1 standard deviation above the mean.
Let <em>X</em> be defined as the scores of students at the National Financial Capability Challenge Exam.
It is provided that the students who score in the top 23% are recognized publicly for their achievement by the Department of the Treasury.
That is, P (X > x) = 0.23.
⇒
⇒
The value of <em>z</em> for this probability value is:
<em>z</em> = 0.74.
*Use a <em>z</em>-table.
Thus, the students have to score 0.74 standard deviations above the mean to be publicly recognized.
Answer:
B) A one-sample t-test for population mean would be used.
Step-by-step explanation:
The complete question is shown in the image below.
The marketing executive is interested in comparing the mean number of sales of this year to that of previous year.
The marketing executive already has the value of mean from previous year and uses a sample to calculate the mean and standard deviation of sales for the current year.
Since, data is being collected for one sample only this limits us to chose between one sample test for mean. So now the possible options are one sample t-test for population mean and one sample t-test for population mean.
If we read the statement we can see that we have the value of sample mean and sample standard deviation. Value of population standard deviation is unknown. In cases where value of population standard deviation is not known and sample standard deviation is given, t-test is used.
Therefore, we can conclude that A one-sample t-test for population mean would be used.
