Question

Based on historical data, your manager believes that 28% of the
company's orders come from first-time customers. A random sample of
175 orders will be used to estimate the proportion of
first-time-customers. What is the probability that the sample
proportion is greater than than 0.22?

Answer = (Enter your answer as a number accurate to 4
decimal places.)

Answer #1

solution:

Given data

population Proportion (P) = 28% = 0.28

Sample size(n) = 175

Consider the sampling distribution of sample proportion follows normal distribution

So,Here Mean (p) = P = 0.28

Standard deviation (p) =

=

= 0.0339

Probability that sample proportion is greater than 0.22 = P( > 0.22)

=

=

= P(Z>-1.77)

= P(Z<1.77) [since , P(Z>-k) = P(Z<k) ]

= 0.9616 [ using standard normal distribution table ]

Probability that sample proportion
is greater than 0.22 = **0.9616**

Based on historical data, your manager believes that 28% of the
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Answer = (Enter your answer as a number accurate to 4 decimal
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Answer = (Enter your answer as a number accurate to 4
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Answer =
(Enter your answer as a number accurate to 4 decimal
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Note: You should carefully round any z-values you calculate to 4
decimal places to match wamap's approach and calculations.
Answer = (Enter your answer as a number accurate to 4
decimal places.)

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Note: You should carefully round any intermediate values you
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Answer = (Enter your answer as a number accurate to 4 decimal
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Note: You should carefully round any z-values you calculate to 4
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Answer = (Enter your answer as a number accurate to 4
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