a. What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?
b. What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?
d. What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?
Lacking data. Need to have some measure of variability to solve.
To answer these questions, we can use the concept of sampling distributions and the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size (typically greater than 30) from a population with any distribution, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.
a. To find the probability that a sample of 50 male graduates will provide a sample mean within $0.50 of the population mean ($21.68), we need to assume that the population standard deviation is known or provided. Let's assume it is known, let's say it's σ.
Since the sample size is large enough (50), we can use the formula for the standard deviation of the sampling distribution, which is σ/√n, where σ is the population standard deviation and n is the sample size.
In this case, we want the sample mean to be within $0.50 of the population mean, so the range of values for the sample mean is $21.68 ± $0.50.
To calculate the probability, we need to convert this range into z-scores. The z-score formula is (X - μ) / (σ/√n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
The z-score for the upper limit of the range is (21.68 + 0.50 - μ) / (σ/√n), and the z-score for the lower limit of the range is (21.68 - 0.50 - μ) / (σ/√n).
Once we have the z-scores, we can use a standard normal distribution table (or calculator) to find the probability associated with those z-scores. The probability will give us the likelihood of obtaining a sample mean within the specified range.
b. To answer the same question for a sample of 50 female graduates with a population mean of $18.80, we follow the same steps as in part a, using the population standard deviation σ and the appropriate z-scores to find the probability.
d. To find the probability that a sample of 120 female graduates will provide a sample mean more than $0.30 below the population mean, we need to calculate the z-score for this scenario.
The z-score formula is (X - μ) / (σ/√n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, we want the sample mean to be more than $0.30 below the population mean, so the range of values for the sample mean is X < $18.80 - $0.30.
By plugging in the values into the z-score formula, we can calculate the appropriate z-score. Once we have the z-score, we can use a standard normal distribution table to find the probability associated with that z-score.
Remember, these calculations depend on the assumption that the population standard deviation is known or provided. If it is not known, you would need to use the sample standard deviation as an approximation or use t-distributions instead.