A population proportion is 0.59. Suppose a random sample of 663 items is sampled

Question

A population proportion is 0.59. Suppose a random sample of 663 items is sampled randomly from this population. Appendix A Statistical Tables a. What is the probability that the sample proportion is greater than 0.60? b. What is the probability that the sample proportion is between 0.54 and 0.64? c. What is the probability that the sample proportion is greater than 0.57? d. What is the probability that the sample proportion is between 0.52 and 0.55? e. What is the probability that the sample proportion is less than 0.50?

Explanation / Answer

Solution :-

Given that population proportion p = 0.59 , n = 663 , q = 1-p = 0.41

a. p^ = 0.60

Z = (p^ - p)/sqrt(p*q/n)

= (0.60 - 0.59)/sqrt(0.59*0.41/663)

= 0.5235

=> P(p > 0.60) = P(Z > 0.5235) = 0.3015

b.
p^ = 0.54

Z = (0.54 - 0.59)/sqrt(0.59*0.41/663) = -2.6176

p = 0.64

Z = (0.64 - 0.59)/sqrt(0.59*0.41/663) = 2.6176

=> P(0.54 < p < 0.64) = P(-2.6176 < Z < 2.6176) = 0.9912

c. p^ = 0.57

Z = (p^ - p)/sqrt(p*q/n)

= (0.57 - 0.59)/sqrt(0.59*0.41/663)

= -1.0471

=> P(p > 0.57) = P(Z > -1.0471) = 0.8531

d. p^ = 0.52

Z = (0.52 - 0.59)/sqrt(0.59*0.41/663) = -3.6647

p = 0.55

Z = (0.55 - 0.59)/sqrt(0.59*0.41/663) = -2.0941

=> P(0.52 < p < 0.55) = P(-3.6647 < Z < -2.0941) = 0.0182

e. p^ = 0.50

Z = (p^ - p)/sqrt(p*q/n)

= (0.50 - 0.59)/sqrt(0.59*0.41/663)

= -4.7117

=> P(p < 0.50) = P(Z < -4.7117) = 0

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