The problem presented involves assessing the appropriateness of using a Normal model to estimate the probability that the sample mean of tips given by customers exceeds $0.32, given specific statistical parameters.
Given data:
- Population mean (\(\mu\)): $0.29
- Population standard deviation (\(\sigma\)): $0.1164
- Sample size (\(n\)): 50
- Sample mean threshold: $0.32
To determine whether it is appropriate to use a Normal model, we apply the Central Limit Theorem (CLT). The CLT suggests that, for a sufficiently large sample size, the distribution of the sample mean \(\bar{x}\) will be approximately Normal, regardless of the population distribution.
1. Calculate the Standard Error of the Mean (SEM):
[tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \][/tex]
Using the provided data:
[tex]\[ \text{SEM} = \frac{0.1164}{\sqrt{50}} \approx 0.0165 \][/tex]
2. Calculate the Z-score:
[tex]\[ Z = \frac{\bar{x} - \mu}{\text{SEM}} \][/tex]
Using the provided data:
[tex]\[ Z = \frac{0.32 - 0.29}{0.0165} \approx 1.82 \][/tex]
3. Determine the Probability:
Using the Z-score, we can find the probability that the sample mean exceeds $0.32 using the cumulative distribution function (CDF) for the standard normal distribution:
[tex]\[ P(\bar{x} > 0.32) = 1 - \Phi(Z) \][/tex]
For \(Z \approx 1.82\):
[tex]\[ P(\bar{x} > 0.32) \approx 1 - \Phi(1.82) \approx 0.0342 \][/tex]
Putting it all together, we find that:
- The calculated standard error is approximately 0.0165.
- The Z-score is approximately 1.82.
- The probability \(P(\bar{x} > 0.32)\) is approximately 0.0342.
Thus, the appropriate choice from the given options should be:
- "Yes, since the sample size 50 is large enough, the sampling distribution for \(\bar{x}\) would be approximately Normal, and \(P(\bar{x}>32)=0.0372\)."
This is the closest to the calculated probability, which we found to be 0.0342 (note that there might be a slight discrepancy due to rounding differences).
So the correct conclusion is that the sampling distribution of the sample mean can be approximated as Normal, and given the calculations, the probability that the sample mean exceeds $0.32 is indeed correctly computed using this approach.
