Answer:
Probability that the sample will have a mean that is greater than $52,000 is 0.0057.
Step-by-step explanation:
We are given that the population mean for income is $50,000, while the population standard deviation is 25,000. 
We select a random sample of 1,000 people.
<em>Let </em> <em> = sample mean</em>
<em> = sample mean</em>
The z-score probability distribution for sample mean is given by;
                Z =  ~ N(0,1)
  ~ N(0,1)
where,  = population mean = $50,000
 = population mean = $50,000
              = population standard deviation = $25,000
 = population standard deviation = $25,000
             n = sample of people = 1,000
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
So, probability that the sample will have a mean that is greater than $52,000 is given by = P( > $52,000)
 > $52,000)
   P( > $52,000) = P(
 > $52,000) = P(  >
 >  ) = P(Z > 2.53) = 1 - P(Z
 ) = P(Z > 2.53) = 1 - P(Z  2.53)
 2.53)
                                                                     = 1 - 0.9943 = 0.0057 
<em>Now, in the z table the P(Z </em> <em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.</em>
<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.</em>
Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.