Question

Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random sample of 1,000 people, what is the probability that sample will have a mean that is greater than $52,000?

Answer 1

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.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

               Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = population mean = $50,000

            $\sigma$ = 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 $52,000 is given by = P($\bar X$ > $52,000)

  P($\bar X$ > $52,000) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } }$ ) = P(Z > 2.53) = 1 - P(Z $\leq$ 2.53)

                                                                    = 1 - 0.9943 = 0.0057

Now, in the z table the P(Z $\leq$ 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.

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.