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A survey of students at a large university found that 82% had purchased textbooks from an off-campus vendor at least once during their college career. If 45 students are randomly sampled, what is the probability that at least 40 students have purchased textbooks from an off-campus vendor at least once during their college career? g

Answer :

Answer:

Probability that at least 40 students have purchased textbooks from an off-campus vendor at least once during their college career is 0.0688.

Step-by-step explanation:

We are given that a survey of students at a large university found that 82% had purchased textbooks from an off-campus vendor at least once during their college career.

Also, 45 students are randomly sampled.

Let [tex]\hat p[/tex] = sample proportion of students who have purchased textbooks from an off-campus vendor at least once during their college career.

The z-score probability distribution for sample proportion is given by;

                               Z = [tex]\frac{\hat p- p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion = [tex]\frac{40}{45}[/tex] = 0.89

p = population proportion of students who had purchased textbooks from an off-campus vendor at least once during their college career = 82%

n = sample of students = 45

Now, probability that at least 40 students have purchased textbooks from an off-campus vendor at least once during their college career is given by = P( [tex]\hat p[/tex] [tex]\geq[/tex] 0.89)

      P( [tex]\hat p[/tex] [tex]\geq[/tex] 0.89) = P( [tex]\frac{\hat p- p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{0.89-0.82}{\sqrt{\frac{0.89(1-0.89)}{45} } }[/tex] ) = P(Z [tex]\geq[/tex] 1.50) = 1 - P(Z < 1.50)

                                                                      = 1 - 0.9332 = 0.0688

The above probability is calculate by looking at the value of x = 1.50 in the z table which ha an area of 0.9332.

Therefore, the required probability is 0.0688.

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