. A recent news report claims that 15% of women are left-handed. Assume that 15%

Question

. A recent news report claims that 15% of women are left-handed. Assume that 15% is the population proportion of left-handed women. A random sample of 150 women is selected to further study this question. A. What is the shape, mean, and standard deviation of the sampling distribution of the sample proportion for a sample of size 150? b. What is the probability that the sample proportion of left-handed women is greater than 20 percent ( 0.20)? c. What is the probability that the sample proportion of left-handed women is within 4% ( 0.04 ) of the population proportion?

Explanation / Answer

n = 150 , p = 0.15 , q = 1 - 0.15 = 0.85

a)

mean = p = 0.15 = 0.15

standard devition = sqrt(p * q / n) = sqrt ( 0.15 * 0.85 /150)

= 0.029

b) P( p > 0.20)

z = ( 0.20 - 0.15) / 0.029

= 1.724

P( p > 0.20) = P( z > 1.724) = 0.0 423

c) p1 = 0.15 - 0.04 = 0.11 , p2 = 0.15 + 0.04 = 0.19

P( 0.11 < p < 0.19)

z = (( 0.11 - 0.15 ) / 0.029 < z < (0.19 - 0.15)/0.029)

= (-1.379 < z < 1.379)

P( 0.11 < p < 0.19) = (-1.379 < z < 1.379) = 0.8322

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