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marta [7]
3 years ago
5

A school offers band and chorus classes. The table shows the percents of the 1200 students in the school who are enrolled in ban

d, chorus, or neither class. How many students are enrolled in both classes?
Mathematics
1 answer:
julsineya [31]3 years ago
5 0

Using Venn probabilities, it is found that 240 students are enrolled in both classes.

<h3>Venn probabilities:</h3>

The events are:

  • Event A: A student is enrolled in band.
  • Event B: A student is enrolled in chorus.

The supposed percentages, which also represents the probabilities involving a single student, are:

  • 50% of the students involved in the band, hence P(A) = 0.5.
  • 40% of the students involved in the chorus, hence P(B) = 0.4.
  • 30% involved in neither, hence 1 - P(A \cup B) = 0.3 \rightarrow P(A \cup B) = 0.7.

The percentage involved in both is:

P(A \cap B) = P(A) + P(B) - P(A \cup B)

Hence:

P(A \cap B) = 0.5 + 0.4 - 0.7 = 0.2

Then, out of 1200 students:

0.2(1200) = 240

240 students are enrolled in both classes.

To learn more about Venn probabilities, you can take a look at brainly.com/question/25698611

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We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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95% confidence interval

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The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

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