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
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation ![\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
.
A pollster draws a sample of 41 people to interview. This means that
.
What is the probability that the sample mean income is less than 42 (thousands of dollars)?
This probability is the pvalue of Z when
.
Due to the Central Limit Theorem, we use s instead of
in the Zscore formula. So
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{42-43}{4.53}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B42-43%7D%7B4.53%7D)
![Z = -0.22](https://tex.z-dn.net/?f=Z%20%3D%20-0.22)
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).