Answer :
Answer:
There is a 41.29% probability that the sample mean income is less than 42 (thousands of dollars).
Step-by-step explanation:
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]\frac{\sigma}{\sqrt{n}}[/tex]
Problems of normally distributed samples can be 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.
In this problem, we have that:
The mean annual income for people in a certain city (in thousands of dollars) is 43, with a standard deviation of 29. This means that [tex]\mu = 43, \sigma = 29[/tex].
A pollster draws a sample of 41 people to interview. This means that [tex]s = \frac{29}\sqrt{41} = 4.53[/tex].
What is the probability that the sample mean income is less than 42 (thousands of dollars)?
This probability is the pvalue of Z when [tex]X = 42[/tex].
Due to the Central Limit Theorem, we use s instead of [tex]\sigma[/tex] in the Zscore formula. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{42-43}{4.53}[/tex]
[tex]Z = -0.22[/tex]
[tex]Z = -0.22[/tex] has a pvalue of 0.4129.
This means that there is a 41.29% probability that the sample mean income is less than 42 (thousands of dollars).