Question 1168824: Based on historical data, your manager believes that 30% of the company's orders come from first-time customers. A random sample of 97 orders will be used to estimate the proportion of first-time-customers.
What is the probability that the sample proportion is greater than than 0.2?
Answer = (Enter your answer as a number accurate to 4 decimal places.)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**1. Define the Parameters**
* Population proportion (p) = 0.30
* Sample size (n) = 97
* Sample proportion (p̂)
* We want to find P(p̂ > 0.2)
**2. Check the Conditions for a Normal Approximation**
To use the normal approximation for the sampling distribution of p̂, we need to check if:
* np ≥ 10
* n(1 - p) ≥ 10
Let's check:
* np = 97 * 0.30 = 29.1 ≥ 10 (condition met)
* n(1 - p) = 97 * 0.70 = 67.9 ≥ 10 (condition met)
Since both conditions are met, we can use the normal approximation.
**3. Calculate the Mean and Standard Deviation of the Sampling Distribution**
* Mean of p̂ (μ_p̂) = p = 0.30
* Standard deviation of p̂ (σ_p̂) = √[p(1 - p) / n]
* σ_p̂ = √[(0.30 * 0.70) / 97]
* σ_p̂ = √(0.21 / 97)
* σ_p̂ = √0.00216494845
* σ_p̂ ≈ 0.04652
**4. Calculate the Z-score**
We want to find P(p̂ > 0.2). To do this, we need to convert 0.2 to a z-score:
* z = (p̂ - μ_p̂) / σ_p̂
* z = (0.2 - 0.30) / 0.04652
* z = -0.10 / 0.04652
* z ≈ -2.15
**5. Find the Probability**
We want to find P(p̂ > 0.2), which is equivalent to P(Z > -2.15).
* Using a standard normal distribution table or calculator, we find:
* P(Z < -2.15) ≈ 0.0158
* Since P(Z > -2.15) = 1-P(Z < -2.15)
* P(Z > -2.15) = 1-0.0158 = 0.9842
**6. Final Answer**
The probability that the sample proportion is greater than 0.2 is approximately 0.9842.
Answer = 0.9842
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