Let's solve this step by step:
1. Identify the given values:
- The sample proportion ([tex]\( \hat{p} \)[/tex]) of customers satisfied with the service is 85% or [tex]\( 0.85 \)[/tex].
- The observed proportion ([tex]\( p \)[/tex]) from Nico's sample is 83% or [tex]\( 0.83 \)[/tex].
- The sample size ([tex]\( n \)[/tex]) is 100.
2. Calculate the standard error of the proportion.
The standard error (SE) for a proportion can be calculated using the formula:
[tex]\[
SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}
\][/tex]
Substituting the given values:
[tex]\[
SE = \sqrt{\frac{0.85 \times (1 - 0.85)}{100}} = \sqrt{\frac{0.85 \times 0.15}{100}} = \sqrt{\frac{0.1275}{100}} = \sqrt{0.001275} \approx 0.0357
\][/tex]
3. Calculate the z-score.
The z-score is calculated using the observed proportion, the sample proportion, and the standard error:
[tex]\[
z = \frac{p - \hat{p}}{SE}
\][/tex]
Substituting the values we have:
[tex]\[
z = \frac{0.83 - 0.85}{0.0357} \approx -0.56
\][/tex]
4. Determine the probability from the z-score.
To find the probability that 83% or more of Nico’s customers are satisfied, we need to find the probability associated with a z-score of -0.56. The z-score represents the number of standard deviations the observed proportion is below the sample proportion.
Using the z-table or standard normal distribution:
- A z-score of -0.56 corresponds to a cumulative probability (to the left) of approximately 0.288 (you can find this in the z-table).
Since we want the probability of 83% or more ranking the service as satisfactory, we use the complement:
[tex]\[
P(Z > -0.56) = 1 - P(Z \leq -0.56) = 1 - 0.288 = 0.712
\][/tex]
Thus, the probability that 83% or more of Nico's customers rank the service as satisfactory is approximately 0.712.
Conclusion:
- Standard error: [tex]\(0.0357\)[/tex]
- Z-score: approximately [tex]\(-0.56\)[/tex]
- Probability [tex]\(83\% \text{ or more}\)[/tex] are satisfied: approximately [tex]\(0.712\)[/tex] or [tex]\(71.2\%\)[/tex]
