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viva [34]
2 years ago
5

In a history class, 30 students were asked about other courses that they were enrolled in: 20 said they were taking math, 22 wer

e taking English and 26 students were in either math or English. a) How many students are taking both math and English b) How many are in neither c) What percent of the students are taking math but not English
Mathematics
1 answer:
zimovet [89]2 years ago
8 0

16 students take both maths and english, 4 students took neither and 15.38% of the students are taking math but not English

<h3>What is probability?</h3>

Probability is the likelihood of occurrence of an event. It is given by:

Probability = number of occurrence / total possible outcomes.

Let x represent the number of students that took both maths and english, hence:

Number of students taking only maths = 20 - x

Number of students taking only english = 22 - x

26 = 20 - x + (22 - x) + x

42 - x = 26

x = 16

b) Those who took neither = 30 - 26 = 4

c) those taking only math = 20 - 16 = 4.

Percentage = (4/26) * 100 = 15.38%

16 students take both maths and english, 4 students took neither and 15.38% of the students are taking math but not English

Find out more on probability at: brainly.com/question/24756209

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 8.4, \sigma = 1.8, n = 40, s = \frac{1.8}{\sqrt{40}} = 0.2846

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This is 1 subtracted by the pvalue of Z when X = 9.1. So

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Z = \frac{9.1 - 8.4}{0.2846}

Z = 2.46

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