**Title: Analyzing Employee Health Care Costs****Introduction:**A college with 2,150 full-time employees is currently analyzing their health care plan. The average out-of-pocket cost for employees is $1,900, with a standard deviation of $250. A random sample of 30 employees has been selected to analyze their out-of-pocket costs.**Problem Statements and Solutions:**a. **Standard Error of the Mean:** - Calculate the standard error of the mean based on the given data. - Formula: SE = σ / √n - **Solution:** (Round to three decimal places)b. **Probability Sample Mean < $1,890:** - Find the probability that the sample mean will be less than $1,890. - **Solution:** (Round to four decimal places)c. **Probability Sample Mean > $1,900:** - Find the probability that the sample mean will be more than $1,900. - **Solution:** (Round to four decimal places)d. **Probability Sample Mean Between $1,865 and $2,005:** - Find the probability that the sample mean will be between $1,865 and $2,005. - **Solution:** (Round to four decimal places)**Conclusion:**This analysis helps in understanding the distribution of out-of-pocket costs among employees and in making informed decisions about health care benefits. Calculating probabilities of different sample mean intervals provides insights into cost fluctuations and potential financial impacts on employees.

a. Define a random variable X, that is the out0of-pocket cost for the employees.According to the given information, X have follows normal distribution with mean µx = $1940 and the standard deviation of σ x = $510. The formula for standard error is σ/√n, where σ is the population standard deviation and n is the sample size. The sample size is n = 30. Hence, the standard error is, 510/√30 = 93.11. Thus, the standard error of the mean is 93.11. b. Consider that the mean and standard deviation of a random variable X are µ and σ, respectively. Thus, the z-score of that random variable X is z = (X – µ)/σ. Central limit theorem: If the population follows normal distribution or the sample size is large enough (>30), then the sampling distribution of the sample mean follows normal distribution with the expectation and standard deviation explained as follows: E(x̅) = µ and SD(x̅) = σ/√n. Here, for the random variable x the population mean is µ, the population standard deviation is σ, and the sample size is n. Denote that x̅ is the sample mean. Now, for sample of 30, the mean of the sampling distribution of the sample mean is, µx̅ = 1940. Now, for sample of 30, the standard deviation of the sampling distribution of the sample mean is, σx̅ = 510/√30 = 93.11283478. Thus, the sampling distribution of x̅ is approximately normal with mean µx̅ = 1940 and standard deviation σx̅ = 93.11283478. The probability that the sample mean will be less than 1890 is, Thus, the probability that the sample mean will be less than 1890 is 0.2956. c. The probability that the sample mean will be more than 1900 is, Thus, the probability that the sample mean will be more than 1900 is 0.6663. d. The probability that the sample mean will be between 1965 and 2005 is, Thus, the probability that the sample mean will be between 1965 and 2005 is 0.1516.