Is 0.3333333 a natural number
2 answers:
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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>
The z-score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean = $50,000
= 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)
P( > $52,000) = P(
>
) = P(Z > 2.53) = 1 - P(Z
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>
Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.