Question

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2.   The president of Doerman Distributors, Inc. believes that 30% of the firm’s orders come from new or first-time customers. A simple random sample of 100 orders is used to estimate the proportion of new or first-time customers. The results of the sample will be used to verify the president’s claim of p = 0.30.

a.   Show whether the study meets the success/failure condition. (1 point)

b.   What is the probability that the sample proportion is less than 0.20 [P(z ≤ 0.20)? (1 point)

c.   What is the probability that the sample proportion is less than 0.40 [P(z ≤ 0.40)? (1 point)

d.   What is the probability that the sample proportion will be between 0.20 and 0.40 [P(0.20 ≤ ≤ 0.40)]? (1 point)

Answer 1

a)

Success / failure conditions are

np = 100 * 0.30 = 30 >= 10 ---> holds.

nq = 100 * (1-0.30) = 70 >= 10 ----> holds.

The study meets success / failure conditions.

b)

Using central limit theorem,

P( < p ) = P( Z < - p / sqrt( p( 1 - p) / n) )

P( < 0.30 ) = P( Z < 0.20 - 0.30 / sqrt( 0.30 * 0.70 / 100) )

= P( Z < -2.1822)

= 0.0145

b)

P( < 0.40) = P( Z < 0.40 - 0.30 / sqrt( 0.30 * 0.70 / 100) )

= P( Z < 2.1822)

= 0.9855

c)

P( 0.20 <= <= 0.40) = P( <= 0.40) - P( <= 0.20)

= P( Z < 0.40 - 0.30 / sqrt( 0.30 * 0.70 / 100) ) + P( Z < 0.20 - 0.30 / sqrt( 0.30 * 0.70 / 100) )

= P( Z < 2.1822) - P( Z < -2.1822)

= 0.9710