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A preliminary survey shows that 34% of college students smoke. In a class of 37 students,...

A preliminary survey shows that 34% of college students smoke. In a class of 37 students, what is the probability that more than 49% of the students smoke? Use Appendix B.1 for the z-values. (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)

Probability ??

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Answer #1

solution:-

given that p = 0.34 , n = 37

standard deviation = sqrt(p*(1-p)/n) = sqrt(0.34*(1-0.34)/37)

   = 0.0754


P(x > 0.49) = P(z > (0.49-0.34)/0.0754)

= P(z > 1.98)

= 1-P(z < 1.98)

= 1 - 0.9761

= 0.0239.

.

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