Chapter 06, Section 6.5, Problem 050 According to a survey, 17% of U.S. adults w

Question

Chapter 06, Section 6.5, Problem 050 According to a survey, 17% of U.S. adults with online services currently read e-books. Assume that this percentage is true for the current population of U.S. adults with online services. Find to 4 decimal places the probability that in a random sample of 700 U.S. adults with online services, the number who read e-books is a. exactly 98. Probability b. at most 104. Probability c. 74 to 103. Probability Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

n = 700

p = 0.17

q = 1 - p = 0.83

Mean = np

= 700x0.17

= 119

Standard deviation = sq root(npq)

= sq root (700x0.17x0.83)

= sq root(98.77)

= 9.94

P(X < A) = P(Z < (A - mean) / standard deviation)

a) P(X = 98) = P(97.5 < X < 98.5)

= P(X < 98.5) - P(X < 97.5)

= P(Z < (98.5 - 119)/9.94) - P(Z < (97.5 - 119)/9.94)

= P(Z < -2.06) - P(Z < -2.16)

= 0.0197 - 0.0154

= 0.0043

b) P(at most 104) = P(X < 104.5) {continuity correction applied}

= P(Z < (104.5 - 119)/9.94)

= P(Z < -1.46)

= 0.0721

c) P(74 to 103) = P(X < 103) - P(X < 74)

= P(Z < (103 - 119)/9.94) - P(Z < (74 - 119)/9.94)

= P(Z < -1.61) - P(Z < -4.53)

= 0.0537 - 0

= 0.0537

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