Respuesta :
Answer:
1. Due to the lower standard deviation, it is more likely to obtain a sample of females within $10,000 of the population mean
2. 15.87% probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
1. In which of the preceding two cases, part a or part b, do we have a higher probability of obtaining a smaple estimate within $10,000 of the population mean? why?
The lower the standard deviation, the less dispersed the values are, meaning it is more likely to find values within a certain threshold of the mean.
So
Due to the lower standard deviation, it is more likely to obtain a sample of females within $10,000 of the population mean.
2. What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean?
We have that:
[tex]\mu = 168000, \sigma = 40000, n = 100, s = \frac{40000}{\sqrt{100}} = 4000[/tex]
This probability is the pvalue of Z when X = 168000 - 4000 = 164000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{164000 - 168000}{4000}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
15.87% probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean
