A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teache
rs who were assigned all curricular resources for use in their classes. On average, when surveyed regarding job satisfaction, teachers give a score of 3.3 out of 5 with a standard deviation of 0.6. When the authors of the study interviewed 40 teachers who supplemented with their own materials, they found 3.5 to be the mean. The authors wanted to know if the group of teachers that could choose supplementary curriculum had a higher level of job satisfaction. They used a significance level of 1%. Which of the following statements is valid based on the results of the test a. The data shows that the authors make a stong proint with the date
b. The data shows that the authors cannot make a determination either way with this data.
c. 70% teachers are not satisfied with their jobs
d. None of the above
There is a 38.97% probability that this student earned an A on the midterm.
Step-by-step explanation:
The first step is that we have to find the percentage of students who got an A on the final exam.
Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.
This means that
Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.
So, the percentage of students who got an A on the final exam is
To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.
The percentage of students who got an A on both tests is:
The probability that the student also earned an A on the midterm is
There is a 38.97% probability that this student earned an A on the midterm.
The simplest fraction for is . Write the upper bound as a fraction with the same denominator:
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Hence the range for would be:
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If the denominator of is also , then the range for its numerator (call it ) would be . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than .
To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)
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At this point, the difference between the numerators is now . That allows a number ( in this case) to fit between the bounds. However, can't be written as finite decimals.
Try multiplying the numerator and the denominator by a different number.
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It is important to note that some expressions for can be simplified. For example, because of the common factor .
So since her balance must be at least 500, her ballance cannot reach 500 so 794-500=294 see how many 25's you can fit into 294 we know that 4 25's =100 and 294 is roughly 300 so 4 times 3=12 we minus one of the 25's because 294 is less than 300 so the innequality is >means more than 794-25x>500