Answer :
Answer:
15.15% probability that the sample mean will be $192,000 or more.
Step-by-step explanation:
To solve this problem, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using 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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 189000, \sigma = 20500, n = 50, s = \frac{20500}{\sqrt{50}} = 2899.14[/tex]
The probability that the sample mean will be $192,000 or more is
This is 1 subtracted by the pvalue of z when X = 192000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{192000 - 189000}{2899.14}[/tex]
[tex]Z = 1.03[/tex]
[tex]Z = 1.03[/tex] has a pvalue of 0.8485.
1-0.8485 = 0.1515
15.15% probability that the sample mean will be $192,000 or more.