To determine the probability that the mean number of daily emergency room admissions for a random sample of 30 days is greater than 70, we can follow these steps:
### Step-by-Step Solution
1. Identify the Given Information:
- Population mean ([tex]\(\mu\)[/tex]): 85
- Population standard deviation ([tex]\(\sigma\)[/tex]): 37
- Sample size ([tex]\(n\)[/tex]): 30
- Mean threshold: 70
2. Calculate the Standard Error of the Mean (SEM):
The standard error of the mean is given by:
[tex]\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\][/tex]
Here, [tex]\(\sigma = 37\)[/tex] and [tex]\(n = 30\)[/tex].
[tex]\[
\text{SEM} = \frac{37}{\sqrt{30}} \approx 6.755
\][/tex]
3. Find the Z-score:
The Z-score tells us how many standard errors the mean threshold is away from the population mean. It is calculated using the formula:
[tex]\[
Z = \frac{\text{Mean threshold} - \mu}{\text{SEM}}
\][/tex]
Substituting the given values:
[tex]\[
Z = \frac{70 - 85}{6.755} \approx -2.220
\][/tex]
4. Determine the Probability Using the Z-score:
Using the Z-score table or a cumulative distribution function (CDF) of the normal distribution, we can find the probability that a Z-score is less than -2.220. The cumulative probability for [tex]\(Z = -2.220\)[/tex] is approximately 0.0132.
5. Calculate the Probability for the Mean Being Greater Than 70:
Since we are looking for the probability that the mean is greater than 70, we need to find the complement of the cumulative probability we found in the previous step:
[tex]\[
\text{Probability (mean > 70)} = 1 - \text{Cumulative probability (Z < -2.220)}
\][/tex]
[tex]\[
\text{Probability} = 1 - 0.0132 \approx 0.9868
\][/tex]
### Conclusion
Therefore, the probability that the mean number of daily emergency room admissions for a random sample of 30 days is greater than 70 is approximately 0.9868 or 98.68%.
### Probability Statement
[tex]\[ P(\bar{X} > 70) = 0.9868 \][/tex]
